Solution

First women choose the chairs from among $1$ to $4$ chairs. i.e., total number of chairs is $4$ . Since, there are two women, so number of arrangements $={ }^{4} P_2$ ways.

Now, men have to choose chairs from remaining $6$ chairs.

Since, there are $3$ men, so number can be arranged in ${ }^{6} P_3$ ways.

$\therefore$ Total number of possible arrangements $={ }^{4} P_2 \times{ }^{6} P_3$

$ \begin{aligned} & =\dfrac{4 !}{4-2 !} \times \dfrac{6 !}{6-3 !} \\ & =\dfrac{4 !}{2 !} \times \dfrac{6 !}{3 !} \\ & =\dfrac{4 \times 3 \times 2 !}{2 !} \times \dfrac{6 \times 5 \times 4 \times 3 !}{3 !} \\ & =4 \times 3 \times 6 \times 5 \times 4=1440 \end{aligned} $

Solution

The letters of the word ‘$\mathrm{RACHIT}$’ in alphabetical order are $\mathrm{A}$, $\mathrm{C}$, $\mathrm{H}$, $\mathrm{I}$, $\mathrm{R}$ and $\mathrm{T}$.

Now, words beginning with $A=5!$

words beginning with $C=5!$

words beginning with $H=5!$

words beginning with $I=5!$

Word beginning with $\mathrm{R}$ i.e., $\mathrm{RACHIT}$ $=1$

$\therefore \quad$ Rank of the word ‘$\mathrm{RACHIT}$’ in dictionary $=5! \times 5! \times 5! \times 5!+1$

$=4 \times 5 ! +1=4 \times 120+1$

$ =480+1=481 $

Solution

Since, candidate cannot attempt more than $5$ questions from either group.

Thus, he is able to attempt minimum two questions from either group.

The number of questions attempted from each group is given in following table

$\begin{array}{|c|c|c|c|c|} \hline \text{Group I} & 5 & 4 & 3 & 2 \\ \hline \text{Group II} & 2 & 3 & 4 & 5 \\ \hline \end{array}$

Since, each group have $6$ questions and total attempted $7$ questions.

$\therefore \ $ Total number of possible ways $={ }^{6} C_5 \times{ }^{6} C_2+{ }^{6} C_4 \times{ }^{6} C_3+{ }^{6} C_3 \times{ }^{6} C_4+{ }^{6} C_2 \times{ }^{6} C_5$

$ \begin{aligned} & =2[{ }^{6} C_5 \times{ }^{6} C_2+{ }^{6} C_4 \times{ }^{6} C_3] \\ \\ & =2[6 \times 15+15 \times 20] \\ \\ & =2[90+300] \\ \\ & =2 \times 390=780 \end{aligned} $

Solution

Total number of points $=18$

Out of which $5$ points are collinear, we get a straight line by joining any two points.

$\therefore \ $ Number of straight line formed by joining the $18$ points taking $2$ at a time $={ }^{18} C_2$

and number of straight line formed by joining $5$ points taking $2$ at a time $={ }^{5} C_2$

But $5$ collinear points, when joined pairwise give only one line.

$\therefore \ $ Required number of straight line $={ }^{18} C_2-{ }^{5} C_2+1$

$ =153-10+1=144 $

Solution

Total number of person $=8$

Number of person to be selected $=6$

It is given that, if $A$ is chosen then, $B$ must be chosen.

Therefore, following cases arise.

Case I When $A$ is chosen, $B$ must be chosen

Number of ways $={ }^{8-2} C_{6-2}={ }^{6} C_4$

Case II When $A$ is not chosen.

Then, $B$ may be chosen.

$\therefore \ $ Number of ways $={ }^{8-1} C_6={ }^{7} C_6$

Hence, required number of ways $={ }^{6} C_4+{ }^{7} C_6$

$ =15+7=22 $

Solution

$\because \ $ Total number of persons $=12$

and number of persons to be selected $=5$

Out of $12$ persons a chairperson is selected $={ }^{12} C_1=12$ ways

Now, remaining $4$ persons are selected out of 11 persons.

$\therefore$ Number of ways $={ }^{11} C_4=330$

$\therefore$ Total number of ways to form a committee of $5$ persons $=12 \times 330=3960$

Solution

There are $26$ English alphabets and $10$ digits ($0$ to $9$).

Since, it is given that each plate contains two different letters followed by three different digits.

$\therefore \ $ Arrangement of 26 letters, taken 2 at a time $={ }^{26} P_2=\dfrac{26 !}{24 !}=26 \times 25=650$

and three-digit number can be formed out of the 10 digits $={ }^{10} P_3=10 \times 9 \times 8=720$ ways

$\therefore$ Total number of licence plates $=650 \times 720=468000$

Solution

It is given that bag contains 5 black and 6 red balls.

So, 2 black balls is selected from 5 black balls in ${ }^{5} C_2$ ways.

and 3 red balls are selected from 6 red balls in ${ }^{6} C_3$ ways.

$\therefore$ Total number of ways in which 2 black and 3 red balls are selected $={ }^{5} C_2 \times{ }^{6} C_3$

$=10 \times 20=200$ ways

Solution

Total number of things $=n$

We have to arrange $r$ things out of $n$ in which three things must occur together.

Therefore, combination of $n$ things taken $r$ at a time in which 3 things always occurs

$ ={ }^{n-3} C_{r-3} $

If three things taken together, then it is considered as 1 group.

Arrangement of these three things $=3!$

Now, $ \ $ we have to arrange $=r-3+1=(r-2)$ objects

$\therefore \ $ Arranged of $(r-2)$ objects $=r-2!$

$\therefore$ Total number of arrangements $={ }^{n-3} C_{r-3} \times (r-2) ! \times 3!$

Solution

Number of letters in the word ‘TRIANGLE’ $=8$, out of which 5 are consonants and 3 are vowels.

If vowels are not together, then we have following arrangement.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{V} & \text{C} & \text{V} & \text{C} & \text{V} & \text{C} & \text{V} & \text{C} & \text{V} & \text{C} & \text{V} \\ \hline \end{array}$

Consonants can be arranged in $=5 !=120$ ways and vowels can occupy at 6 places.

The 3 vowels can be arranged at 6 place in ${ }^{6} P_3$ ways $=\dfrac{6 !}{6-3 !}=\dfrac{6 !}{3 !}$

$ =\dfrac{6 \times 5 \times 4 \times 3 !}{3 !}=120 $

Total number of arrangement $=120 \times 120=14400$

Solution

We know that a number is divisible by 5 , If at the units place of the number is 0 or 5 .

We have to form 4 -digit number which is greater than 6000 and less than 7000 . So, unit digit can be filled in 2 ways.

Since, repeatition is not allowed. Therefore, tens place can be filled in 7 ways, similarily hundreds place can be filled in 8 ways.

But we have to form a number greater than 6000 and less than 7000 .

Hence, thousand place can be filled in only 1 ways.

$\begin{array}{|c|c|c|c|c|} \hline \text{Place} & \text{Th} & \text{H} & \text{T} & \text{U} \\ \hline \text{Ways} & 1 & 8 & 7 & 2 \\ \hline \end{array} $

$\therefore \ $ Total number of integers $=1 \times 8 \times 7 \times 2$

$ \hspace{3.7cm}=14 \times 8=112 $

Solution

Given that, $P_1, P_2, \ldots, P_{10}$, are 10 persons, out of which 5 persons are to be arranged but $P_1$ must occur whereas $P_4$ and $P_5$ never occur.

$\therefore$ Selection depends on only $10-3=7$ persons

As, we have already occur $P_1$, Therefore, we have to select only 4 persons out of 7 .

Number of selection $={ }^{7} C_4=\dfrac{7 !}{4 !(7-4) !}=\dfrac{7 !}{4 ! 3 !}=\dfrac{5040}{24 \times 6}=35$

$\therefore \ $ Required number of arrangement of 5 persons $=35 \times 5 !=35 \times 120=4200$

Thinking Process

The number of ways in which the hall can be illuminated is equivalent to the number of selections of one or more things out of $n$ different things is

$ { }^{n} C_1+{ }^{n} C_2+{ }^{n} C_3+\ldots+{ }^{n} C_{n}=2^{n}-1 $

Solution

Since at least one lamp is to be kept switched on.

$\therefore \ $ Total number of ways $={ }^{10} C_1+{ }^{10} C_2+{ }^{10} C_3+{ }^{10} C_4+{ }^{10} C_5+{ }^{10} C_6+\ldots+{ }^{10} C_{10}$

$ \begin{aligned} & =2^{10}-1 \quad[\because{ }^{n} C_0+{ }^{n} C_1+{ }^{n} C_2+\ldots=2^{n}] \\ & =1024-1=1023 \end{aligned} $

Solution

There are 2 white, three black and four red balls.

We have to draw 3 balls, out of these 9 balls in which atleast one black ball is included.

Hence, we can select the balls in the following ways.

$\begin{array}{|c|c|c|c|} \hline \text{Black balls} & 1 & 2 & 3 \\ \hline \text{Other than black} & 2 & 1 & 0 \\ \hline \end{array}$

$\therefore \ $ Required number of selections $={ }^{3} C_1 \times{ }^{6} C_2+{ }^{3} C_2 \times{ }^{6} C_1+{ }^{3} C_3 \times{ }^{6} C_0$

$ \begin{aligned} & =3 \times 15+3 \times 6+1 \\ & =45+18+1=64 \end{aligned} $

Solution

Given, ${ }^n C_{r-1} = 36 \quad \ldots (i)$

$\Rightarrow { }^n C_r = 84 \quad \ldots (ii)$

$\Rightarrow { }^n C_{r+1} = 126 \quad \ldots (iii)$

On dividing Eq. (i) by Eq. (ii), we get

$\dfrac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}} =\dfrac{36}{84} $

$\dfrac{n!}{(r-1)![n-(r-1)!]} \cdot \dfrac{r!(n-r)!}{n!} = \dfrac{3}{7} \quad $ $\Big[\because{ }^{n} C_{r}=\dfrac{n !}{(n-r ! r !)} \Big]$

$\Rightarrow \ \dfrac{1}{(r-1)!(n-r+)!} \cdot \dfrac{r(r-1)!(n-r)!}{1} = \dfrac{3}{7} \quad \Big[ \because n !=n(n-1) !\Big]$

$\Rightarrow \ \dfrac{r(n-r)!}{(n-r+1)(n-r)!} = \dfrac{3}{7}$

$\Rightarrow \ \dfrac{r}{n-r+1} = \dfrac{3}{7}$

$\Rightarrow \ 7r = 3n-3r+3$

$\Rightarrow \ 10r-3n = 3 \quad \ldots (iv)$

On dividing Eq. (ii) by Eq. (iii), we get

$\dfrac{{}^nC_r}{{}^nC_{r+1}} = \dfrac{84}{126}$

$\Rightarrow \ \dfrac{n!}{r!(n-r)!} \cdot \dfrac{(r+1)![n-(r+1)]!}{n!} = \dfrac{84}{126} =\dfrac{14}{21}$

$\Rightarrow \ \dfrac{1}{r!(n-r)!} \cdot \dfrac{(r+1)r!(n-r-1)!}{1} = \dfrac{14}{21}$

$\Rightarrow \ \dfrac{1}{(n-r)(n-r-1)!} \cdot \dfrac{(r+1)(n-r-1)!}{1} = \dfrac{2}{3}$

$\Rightarrow \ \dfrac{n-r}{r+1} = \dfrac{3}{2}$

$\Rightarrow \ 2n-2r = 3r+1$

$\Rightarrow \ 2n-5r= 3 \quad \ldots (v)$

Multiplying by 2 with Eq. (iv) and 3 with Eq. (v), we get

$20r-6n = 6$

and $6n-15r = 9$

Adding these two equations

$5r=15$

$\Rightarrow \ r=3$

$\therefore \ {}^rC_2 = {}^3C_2 = \dfrac{3!}{2! \times 1!} = 3$

Solution

Here, we have to find the number of integers greater than 7000 with the digits $3,5,7,8$ and 9 . So, with these digits we can make maximum five-digit number because repeatition is not allowed.

Now, all the five-digit numbers are greater than 7000 .

Number of ways of forming 5-digit number $=5 \times 4 \times 3 \times 2 \times 1=120$

and all the four-digit numbers greater than 7000 can be formed in following manner.

Thousand place can be filled in 3 ways. Hundred place can be filled in 4 ways. Tenth place can be filled in 3 ways. Units place can be filled in 2 ways.

Thus, we have total number of 4-digit number $=3 \times 4 \times 3 \times 2=72$

$\therefore \ $ Total number of integers $ = 120+72=192$

Solution

It is given that no two lines are parallel means all line are intersecting and no three lines are concurrent means three lines intersect at a point.

Since, we know that for one point of intersection, we required two lines.

$\therefore \quad$ Number of point of intersection $={ }^{20} C_2=\dfrac{20 !}{2 ! 18 !}=\dfrac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}$

$ =\dfrac{20 \times 19}{2}=19 \times 10=190 $

Solution

If first two digit is 41 , the remaining 4 digits can be arranged in

$ \begin{aligned} & ={ }^{8} P_4=\dfrac{8 !}{8-4 !}=\dfrac{8 !}{4 !} \\ & =\dfrac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !} \\ & =8 \times 7 \times 6 \times 5=1680 \end{aligned} $

Similarly, if first two digit is $42,46,62$, or 64 , the remaining 4 digits can be arranged in ${ }^{8} P_4$ ways i.e., 1680 ways.

$\therefore$ Total number of telephone numbers have all six digits distinct $=5 \times 1680=8400$

Solution

It is given that 2 questions are compulsory out of 5 questions.

So, these two questions are always included in the selection.

We know that, the selection of $n$ distinct objects taken $r$ at a time in which $p$ objects are always included in ${ }^{n-p} C_{r-p}$ ways.

$\therefore \quad$ Total number of ways $={ }^{5-2} C_{4-2}={ }^{3} C_2$

$ =\dfrac{3 !}{2 ! 1 !}=\dfrac{3 \times 2 !}{2 !}=3 $

Solution

Let the convex polygon has $n$ sides.

No. of lines possible with vertices $= {}^nC_2$

$\therefore \quad$ Number of diagonals $={ }^{n} C_2-n$

According to the question,

${ }^{n} C_2-n =44 $

$\dfrac{n!}{2!(n-2)!} - n = 44$

$\Rightarrow \ \dfrac{n(n-1)(n-2)!}{2(n-2)!} - n = 44$

$\Rightarrow \dfrac{n(n-1)}{2}-n =44 $

$\Rightarrow n \Big[\dfrac{n-1}{2} - 1\Big] = 44 $

$\Rrightarrow n\Big (\dfrac{n-1-2}{2} \Big) = 44$

$\Rightarrow n(n-3) =44 \times 2 \Rightarrow n(n-3)=88 $

$\Rightarrow n^{2}-3 n-88 =0$

$\Rightarrow \ n^2-11n+8n-88 = 0$

$\Rightarrow(n-11)(n+8)=0 $

$\Rightarrow n =11,-8 \quad \quad [\because n \neq -8] $

$\therefore n =11 $

Therefore, number of sides $=11$

Solution

It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.

So, number of ways forming three groups each of size ’ 6 ’ $ ={ }^{18} C_{6}{ }^{12} C_{6}{ }^{6} C_{6}=\dfrac{18!}{6! \times 6! \times 6!} $

But in this counting the order in which three particular groups are formed is also counted.

Clearly, when we form groups the order must not be counted.

Now three particular groups can occur in $3!$ ways when order is counted.

$ \text { So, actual number of ways }=\dfrac{18!}{6! \times 6! \times 6! \times 3!} = \dfrac{18!}{(6!)^3 \times 3!} $

Solution

Total number of marbles $=6$ white $+5$ red $=11$ marbles

(a) If they can be of any colour means we have to select $4$ marbles out of $11$.

$\therefore \ $ Required number of ways $={ }^{11} C_4$

(b) If two must be white, then selection will be ${ }^{6} C_2$ and two must be red, then selection will be ${ }^{5} C_2$.

$\therefore \ $ Required number of ways $={ }^{6} C_2 \times{ }^{5} C_2$

(c) If they all must be of same colour, then selection of $4$ white marbles out of $6={ }^{6} C_4$

and selection of $4$ red marble out of $5={ }^{5} C_4$

$\therefore \ $ Required number of ways $={ }^{6} C_4+{ }^{5} C_4$

Solution

Total number of players $=16$

We have to select a team of $11$ players

(i) include $2$ particular players $={ }^{16-2} C_{11-2}={ }^{14} C_9$

[since, selection of $n$ objects taken $r$ at a time in which $p$ objects are always included is $ \ { }^{n-p} C_{r-p}$ ]

(ii) Exclude $2$ particular players $={ }^{16-2} C_{11}={ }^{14} C_{11}$

[since, selection of $n$ objects taken $r$ at a time in which $p$ objects are never included is $ \ { }^{n-p} C_{r}$ ]

Solution

Total students in each class $=20$

We have to selects atleast $5$ students from each class.

Hence, selection of sport team of $11$ students from each class is given in following table

$\begin{array}{|c|c|c|} \hline \text{Class XI} & 5 & 6 \\ \hline \text{Class XII} & 6 & 5 \\ \hline \end{array}$

$\therefore \ $ Total number of ways of selecting a team of $11$ players $={ }^{20} C_5 \times{ }^{20} C_6+{ }^{20} C_6 \times{ }^{20} C_5$

$ =2 \times{ }^{20} C_5 \times{ }^{20} C_6 $

Solution

Number of girls $=4$ and Number of boys $=7$

We have to select a team of $5$ members provided that

(i) team having no girls.

$\therefore \ $ Required selection $={ }^{7} C_5=\dfrac{7 !}{5 ! 2 !}=\dfrac{7 \times 6}{2}=21$

(ii) atleast one boy and one girl

$\therefore \ $ Required selection $={ }^{7} C_1 \times{ }^{4} C_4+{ }^{7} C_2 \times{ }^{4} C_3+{ }^{7} C_3 \times{ }^{4} C_2+{ }^{7} C_4 \times{ }^{4} C_1$

$ \begin{aligned} & =7 \times 1+21 \times 4+35 \times 6+35 \times 4 \\ & =7+84+210+140=441 \end{aligned} $

(iii) when atleast three girls are included $={ }^{4} C_3 \times{ }^{7} C_2+{ }^{4} C_4 \times{ }^{7} C_1$

$ =4 \times 21+7=84+7=91 $

Solution

Option (a) Given that,

$ { }^{n} C_{12}={ }^{n} C_8 $

$ \begin{aligned} & \Rightarrow \ { }^{n} C_{n-12}={ }^{n} C_8 \quad \quad {[\because{ }^{n} C_{r}={ }^{n} C_{n-r}]} \\ \\ & \Rightarrow \ n-12=8 \\ \\ & \Rightarrow \ n=12+8=20 \\ \\ \end{aligned} $

• Option (b) 12: If $ n = 12 $, then ${ }^{12} C_{12} \neq { }^{12} C_8$. Specifically, ${ }^{12} C_{12} = 1$ and ${ }^{12} C_8 = 495$, so they are not equal.

• Option (c) 6: If $ n = 6 $, then ${ }^{6} C_{12} $ is not defined because the value of $ r $ in ${ }^{n} C_{r} $ cannot be greater than $ n $.

• Option (d) 30: If $ n = 30 $, then ${ }^{30} C_{12} \neq { }^{30} C_8$. Specifically, ${ }^{30} C_{12} \approx 86493225$ and ${ }^{30} C_8 \approx 5852925$, so they are not equal.

Solution

Option (b) Number of outcomes when tossing a coin $1$ times $=2$ (head or tail)

$\therefore \ $ Total possible outcomes when a coin tossed $6$ times $=2^{6}=64 \quad \quad [\because 2^{n}$ or $n$ time tossed coin $]$

• Option (a) $36$: This is incorrect because the number of possible outcomes when a coin is tossed $6$ times is calculated as $2^6$, not $6 \times 6$. The value $36$ would be relevant if each toss had $6$ possible outcomes, which is not the case for a coin toss.

• Option (c) $12$: This is incorrect because the number of possible outcomes when a coin is tossed $6$ times is calculated as $2^6$, not $2 \times 6$. The value $12$ would be relevant if there were $2$ outcomes per toss and only $6$ total outcomes, which is not the case for multiple coin tosses.

• Option (d) $32$: This is incorrect because the number of possible outcomes when a coin is tossed $6$ times is calculated as $2^6$, which equals $64$, not $32$. The value $32$ would be relevant if the coin were tossed $5$ times, as $2^5 = 32$.

Solution

Option (c) Given, digits $2, 3, 4$ and $7$, we have to form four-digit numbers using these digits.

$\therefore \ $ Required number of ways $={ }^{4} P_4=4 !=4 \times 3 \times 2 !=24$

• Option (a) $120$: is incorrect because it represents the total number of permutations of $5$ digits ($2, 3, 4, 7$, and an additional digit), which is $5! = 120$. However, we are only using $4$ digits.

• Option (b) $96$: is incorrect because it does not correspond to any standard permutation or combination calculation for $4$ digits out of $4$. It might be a result of a miscalculation or misunderstanding of the problem.

• Option (d) $100$: is incorrect because it does not align with the factorial calculation for $4$ digits. There is no mathematical basis for arriving at $100$ permutations using $4$ unique digits.

Solution

Option (b) If we fixed $3$ at units place.

Total possible number is $3 !$ i.e., $6$ .

Sum of the digits in unit place of all these numbers $=3 ! \times 3$

Similarly, if we fixed $4,5$ and $6$ at units place, in each case total possible numbers are $3 !$.

Required sum of unit digits of all such numbers $=(3+4+5+6) \times 3!$

$ =18 \times 3 !=18 \times 6=108 $

• Option (a) $432$: This option is incorrect because it overestimates the sum of the digits in the unit place. The correct calculation involves multiplying the sum of the digits ($3, 4, 5$, and $6$) by the factorial of the remaining digits $(3!)$, which results is $108$, not $432$.

• Option (c) $36$: This option is incorrect because it underestimates the sum of the digits in the unit place. The correct calculation involves multiplying the sum of the digits ($3, 4, 5$, and $6$) by the factorial of the remaining digits $(3!)$, which results is $108$, not $ 36 $.

• Option (d) $18$: This option is incorrect because it significantly underestimates the sum of the digits in the unit place. The correct calculation involves multiplying the sum of the digits ($3, 4, 5$, and $6$) by the factorial of the remaining digits $(3!)$, which results is $108$, not $18$.

Solution

Option (c) Given that, total number of vowels $=4$

and total number of consonants $=5$

Number ways of selcting $2$ vowels and $3$ consonants

$ \begin{aligned} & ={ }^{4} C_2 \times{ }^{5} C_3=\dfrac{4 !}{2 ! 2 !} \times \dfrac{5 !}{3 ! 2 !} \\ \\ & =\dfrac{4 \times 3 !}{2 ! 2 !} \times \dfrac{5 \times 4 \times 3 \times 2 !}{3 ! \times 2 !}=\dfrac{4 \times 5 \times 4 \times 3}{4} \\ \\ & =5 \times 4 \times 3=60 \end{aligned} $

Each of these $5$ letters can be arranged in $5!$ ways i.e., $120$.

So, total number of words $=60 \times 120=7200$

• Option (a) $60$: This option is incorrect because it only accounts for the number of ways to choose the $2$ vowels and $3$ consonants, but it does not consider the different permutations of these $5$ letters to form distinct words. The correct calculation should include the arrangement of these letters, which is done by multiplying by $5!$.

• Option (b) $120$: This option is incorrect because it only considers the number of ways to arrange the $5$ letters $(5!)$, but it does not account for the initial selection of $2$ vowels from 4 and 3 consonants from $5$. The correct calculation should include both the selection and the arrangement.

• Option (d) $720$: This option is incorrect because it underestimates the total number of words. It might be considering only a partial calculation, such as the arrangement of the letters $(5!)$ or a combination of selections, but it does not correctly combine both the selection of vowels and consonants and their arrangement. The correct total should be $7200$.

Solution

Option (a) We know that, a number is divisible by $3$ , when sum of digits in the number must be divisible by $3$ .

So, if we consider the digits $0,1,2,4,5$, then $(0+1+2+4+5)=12$

We see that, sum is divisible by $3$ . Therefore, five-digit numbers using the digit

$0,1,2,4,5=4 \times 4 \times 3 \times 2 \times 1=96$

$\begin{array}{|c|c|c|c|c|c|} \hline \text{Place} & \text{TTh} & \text{Th} & \text{H} & \text{T} & \text{U} \\ \hline \text{Ways} & 4 & 4 & 3 & 2 & 1 \\ \hline \end{array}$

And if we consider the digit $1,2,3,4,5$, then $(1+2+3+4+5=15)$

This sum is also divisible by $3$ .

So, five-digit number can be formed using the digit $1, 2, 3, 4, 5$ in $5 !$ ways.

Total number of ways $=96+5 !=96+120=216$

• Option (b) $600$: This option is incorrect because it overestimates the number of valid five-digit numbers. The correct calculation involves ensuring that the sum of the digits is divisible by $3$ and then counting the permutations. The correct total is $216$, not $600$.

• Option (c) $240$: This option is incorrect because it underestimates the number of valid five-digit numbers. The correct total number of ways to form such numbers, considering the sum of the digits must be divisible by $3$, is $216$, not $240$.

• Option (d) $3125$: This option is incorrect because it vastly overestimates the number of valid five-digit numbers. It likely does not take into account the restriction that the sum of the digits must be divisible by $3$ and the permutations of the digits without repetition. The correct total is $216$, not $3125$.

Solution

Option (b) Let the total number of person in the room is $n$. We know that, two person form $1$ hand shaken.

$\therefore \ $ Required number of hand shakes $={ }^{n} C_2=\dfrac{n !}{2 !(n-2) !}=\dfrac{n(n-1)}{2}$

According to the question, $ \ \dfrac{n(n-1)}{2}=66$

$\Rightarrow \quad n(n-1)=132$

$\Rightarrow \quad n^{2}-n-132=0$

$\Rightarrow \quad(n-12)(n+11)=0$

$\Rightarrow \quad n=12,-11$

Since the number of people cannot be negative, we discard $n=-11$.

$\therefore \quad n=12$

• Option (a) $11$: If there were $11$ people in the room, the number of handshakes would be calculated as $\dfrac{11 \times 10}{2} = 55$. This does not match the given total of $66$ handshakes.

• Option (c) $13$: If there were $13$ people in the room, the number of handshakes would be calculated as $\dfrac{13 \times 12}{2} = 78$. This does not match the given total of $66$ handshakes.

• Option (d) $14$: If there were $14$ people in the room, the number of handshakes would be calculated as $\dfrac{14 \times 13}{2} = 91$. This does not match the given total of $66$ handshakes.

Solution

Option (d) Total number of triangles formed from $12$ points taking $3$ at a time $={ }^{12} C_3$

But out of $12$ points $7$ are collinear. So, these $7$ points constitute a straight line mean no triangle is formed by joining these $7$ points.

$\therefore \ $ Required number of triangles $={ }^{12} C_3-{ }^{7} C_3=220-35=185$

• Option (a) $105$: This option is incorrect because this number is not correct based on our calculations.

• Option (b) $15$: This option is incorrect because this number is significantly lower than expected and incorrect.

• Option (c) $175$: This option is incorrect because this number is close but still less than the calculated result.

Solution

Option (b) To form parallelogram we required a pair of line from a set of $4$ lines and another pair of line from another set of $3$ lines.

$\therefore \ $ Required number of parallelograms $={ }^{4} C_2 \times{ }^{3} C_2=6 \times 3=18$

• Option (a) $6$: This is incorrect because it only considers the combinations from one set of lines.

• Option (c) $12$: This is incorrect because it miscalculates the total number of parallelograms.

• Option (d) $9$: This is incorrect because it underestimates the number of possible parallelograms.

Solution

Option (c) Total number of players $=22$

We have to select a team of $11$ players. Selection of $11$ players when $2$ of them is always included and $4$ are never included.

Total number of players $=22-2-4=16$

$\therefore \ $ Required number of selections $={ }^{16} C_9$

• Option (a) ${ }^{16} C_{11}$: This option is incorrect because it represents the number of ways to select $11$ players from $16$ players. However, we need to select only $9$ players from the remaining $16$ players after including $2$ specific players and excluding $4$ specific players.

• Option (b) ${ }^{16} C_5$: This option is incorrect because it represents the number of ways to select $5$ players from $16$ players. We need to select $9$ players from the remaining $16$ players after including $2$ specific players and excluding $4$ specific players.

• Option (d) ${ }^{20} C_9$: This option is incorrect because it represents the number of ways to select $9$ players from $20$ players. However, after including $2$ specific players and excluding $4$ specific players, we are left with $16$ players, not $20$. Therefore, we need to select $9$ players from these $16$ players.

Solution

Option (d) If all the digits repeated, then number of $5$ digit telephone numbers can be formed in $10^{5}$ ways and if no digit repeated, then $5$ -digit telephone numbers can be formed in ${ }^{10} P_5$ ways.

$\therefore \ $ Required number of ways $=10^{5}-{ }^{10} P_5=100000-\dfrac{10 !}{5 !}$

$ \begin{aligned} & =100000-10 \times 9 \times 8 \times 7 \times 6 \\ \\ & =100000-30240=69760 \end{aligned} $

• Option (a) $90,000$: This option is incorrect because it represents the total number of $5-$ digit telephone numbers that can be formed if we exclude the case where all digits are unique. However, this is not the correct calculation for the number of $5-$ digit telephone numbers with at least one repeated digit. The correct calculation involves subtracting the number of $5-$ digit numbers with all unique digits from the total number of $5-$ digit numbers.

• Option (b) $10,000$: This option is incorrect because it represents the total number of $5-$ digit telephone numbers that can be formed if all digits are the same. This is not relevant to the problem, which asks for the number of $5-$ digit telephone numbers with at least one repeated digit.

• Option (c) $30,240$: This option is incorrect because it represents the number of $5-$ digit telephone numbers with all unique digits. The problem asks for the number of $5-$ digit telephone numbers with at least one repeated digit, which is the total number of $5-$ digit numbers minus the number of $5-$ digit numbers with all unique digits.

Solution

Option (a) $\because \ $ Number of men $=4$

and number of women $=6$

It is given that committee includes two men and exactly twice as many women as men.

Thus, possible selection is given in following table

$\begin{array}{|c|c|} \hline \text{Men} & \text{Women} \\ \hline 2 & 4 \\ \hline 3 & 6 \\ \hline \end{array}$

Required number of committee formed $={ }^{4} C_2 \times{ }^{6} C_4+{ }^{4} C_3 \times{ }^{6} C_6$

$ =6 \times 15+4 \times 1=94 $

• Option (b) $126$: is incorrect because it does not match the calculated number of ways to form the committee, which is $94$. The calculation for the number of ways to choose the committee is based on the combinations of men and women as specified, and $126$ is not the result from these combinations.

• Option (c) $128$: is incorrect because it also does not match the calculated number of ways to form the committee, which is $94$. The calculation involves specific combinations of men and women, and $128$ is not the result of these combinations.

• Option (d) None of these: is incorrect because the correct number of ways to form the committee, as calculated, is $94$, which matches option (a). Therefore, “None of these” is not the correct answer.

Solution

Option (c) We know that the total number of digits $ =10 $

i.e. $ \ 0,1,2,3,4,5,6,7,8,9 $

Out of which $0$ can’t be placed in the first place.

Therefore, first place can be filled in $9$ ways.

The rest $9$ blanks with $9$ digits in $9!$ ways

$\therefore \ $ Required number of ways $=9 \times 9!$

• Option (a) $10!$: is incorrect because it represents the total number of permutations of $10$ digits, but we are only interested in 9-digit numbers. Additionally, it does not account for the restriction that $0$ cannot be the first digit.

• Option (b) $9!$: is incorrect because it represents the total number of permutations of $9$ digits without considering the restriction that $0$ cannot be the first digit. It also does not account for the fact that the first digit has $9$ possible choices (1-9).

• Option (d) $10 \times 10!$: is incorrect because it incorrectly suggests that there are $10$ choices for each of the $10$ positions, which is not the case. It also does not account for the restriction that $0$ cannot be the first digit and the fact that we are dealing with 9-digit numbers, not 10-digit numbers.

Solution

Option (b) Total number of letters in the word ‘ARTICLE’ is $7$, out of which $\mathrm{A, E, I}$ are vowels and $\mathrm{R, T, C, }$ are consonants.

Since, it is given that vowels occupy even place, therefore the arrangement of vowel, consonant can be understand with the help of following diagram.

$ \begin{array}{|c|c|c|c|c|c|c|} \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \end{array} $

Now, vowels can be placed at $2$, $4$ and $6$ th position.

Therefore, number of arrangement $={ }^{3} P_3=3 ! =6$ ways

and consonants can be placed at $1$, $3$, $5$ and $7$ th position.

Therefore, number of arrangement $={ }^{4} P_4=4 ! =24$

$\therefore \ $ Total number of words $=6 \times 24=144$

• Option (a) $1440$: This option is incorrect because it overestimates the number of possible arrangements. The calculation for the number of ways to arrange the vowels and consonants separately and then combine them does not result in such a high number. The correct calculation shows that the total number of words is $144$, not $1440$.

• Option (c) $7!$: This option is incorrect because it represents the total number of permutations of all $7$ letters without any restrictions. The problem specifically requires that vowels occupy the even positions, which imposes additional constraints that reduce the number of valid permutations.

• Option (d) ${ }^{4} C_4 \times{ }^{3} C_3$: This option is incorrect because it uses combinations instead of permutations. The problem requires arranging the letters in specific positions, which is a permutation problem, not a combination problem. The correct approach involves calculating the permutations of vowels and consonants separately and then multiplying them.

Solution

Option (b)

At least one green dye can be chosen in

$ { }^{5} C_{1}+{ }^{5} C_{2}+{ }^{5} C_{3}+{ }^{5} C_{4}+{ }^{5} C_{5}=2^{5}-1 $ ways.

At least one blue dye can be chosen in

$ { }^{4} C_{1}+{ }^{4} C_{2}+{ }^{4} C_{3}+{ }^{4} C_{4}=2^{4}-1 $ ways.

Any number of red dyes can be chosen in

$ { }^{3} C_{0}+{ }^{3} C_{1}+{ }^{3} C_{2}+{ }^{3} C_{3}=2^{3} $ ways.

Then, total number of selection $=(2^{5}-1)(2^{4}-1) \times 2^{3}=3720$

• Option (a) $ 3600 $: This option is incorrect because it does not account for the correct calculation of the total number of combinations. The correct calculation involves subtracting 1 from the total possible combinations of green and blue dyes to ensure at least one of each is chosen, which results in $3720$, not $ 3600 $.

• Option (c) $ 3800 $: This option is incorrect because it overestimates the number of combinations. The correct calculation, which ensures at least one green and one blue dye is chosen, results in $3720$, not $3800$.

• Option (d) $ 3600 $: This option is incorrect for the same reason as option (a). It does not correctly account for the subtraction needed to ensure at least one green and one blue dye is chosen, leading to an incorrect total of $3600$ instead of the correct $3720$.

Solution

Given that, ${ }^{n} P_{r}=840$ and ${ }^{n} C_{r}=35$

$\because \ { }^{n} P_{r}={ }^{n} C_{r} \cdot r !$

$\Rightarrow \ 840=35 \times r ! $

$\Rightarrow \ r !=\dfrac{840}{35}=24 $

$\Rightarrow \ r !=4 \times 3 \times 2 \times 1 $

$\Rightarrow \ r !=4 ! $

$\therefore \ r=4$

Solution

${ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7={ }^{15} C_{15-8}+{ }^{15} C_{15-9}-{ }^{15} C_6-{ }^{15} C_7 \quad[\because{ }^{n} C_{r}={ }^{n} C_{n-r}]$

$ ={ }^{15} C_7+{ }^{15} C_6-{ }^{15} C_6-{ }^{15} C_7 $

$=0$

Solution

Number of permutations of $n$ different things taken $r$ at a time when repetition is allowed $=n^{r}$

Solution

Total number of letters in the word ‘INTERMEDIATE’ $=12$ out of which $6$ are consonants and $6$ are vowels.

The arrangement of these $12$ alphabets in which two vowels never come together can be understand with the help of follow manner.

$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{V} & \text{C} & \text{V} & \text{C} & \text{V} & \text{C} & \text{V} & \text{C} & \text{V} & \text{C} & \text{V} & \text{C} & \text{V} \\ \hline \end{array} $

$6$ consonants out of which $2$ are alike can be placed in $\dfrac{6 !}{2 !}$ ways and $6$ vowels, out of which $3$ E’s alike and $2$ I’s are alike can be arranged at seven place in ${ }^{7} P_6 \times \dfrac{1}{3 !} \times \dfrac{1}{2 !}$ ways.

$\therefore \ $ Total number of words $=\dfrac{6 !}{2 !} \times{ }^{7} P_6 \times \dfrac{1}{3 !} \times \dfrac{1}{2 !}=151200$

Solution

Required number of ways $={ }^{5} C_2 \times{ }^{7} C_1+{ }^{5} C_3 \quad $ [since, at least two red]

$ \begin{aligned} & =10 \times 7+10 \\ & =70+10=80 \end{aligned} $

Solution

Among the digits $0,1,2,3,4,5,6,7,8,9$ clearly $1,3,5,7$ and $9$ are odd.

$\therefore \ $ Number of six-digit numbers $=5 \times 5 \times 5 \times 5 \times 5 \times 5=5^{6}$

Solution

Let the number of team participating in championship be $n$.

Since, it is given that every two teams played one match with each other.

$\therefore \ $ Total match played $={ }^{n} C_2$

According to the question,

${ }^{n} C_2 =153 $

$\Rightarrow \ \dfrac{n(n-1)}{2} =153 $

$\Rightarrow \ n^{2}-n =306 $

$\Rightarrow \ n^{2}-n-306 =0 $

$\Rightarrow \ (n-18)(n+17) =0 $

$\Rightarrow \ n =18,-17 $

Since, the number of teams cannot be negative.

$\therefore \ n =18 $

Solution

The arrangement can be understand with the help of following figure.

$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \ - & + & - & + & - & + & - & + & - & + & - & + & - \\ \hline \end{array} $

Thus, ‘$ +$ ’ sign can be arranged in $1$ way because all are identical and $4$ negative signs can be arranged at $7$ places in ${ }^{7} C_4$ ways.

$\therefore \ $ total number of ways $={ }^{7} C_4 \times 1$

$ \begin{aligned} & =\dfrac{7 !}{4 ! 3 !}=\dfrac{7 \times 6 \times 5 \times 4 !}{3 ! \times 4 !} \\ & =\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}=35 \text { ways } \end{aligned} $

Solution

$\because \ $ Total number of men $=10$

and total number of women $=7$

We have to form a committee containing atleast $3$ men and $2$ women.

Number of ways $={ }^{10} C_3 \times {}^7C_3+{ }^{10} C_4 \times {}^7C_2$

If two particular women to be always there,

Number of ways $={ }^{10} C_4 \times{ }^{5} C_0+{ }^{10} C_3 \times{ }^{5} C_1$

$\therefore \ $ Total number of committee when two particular women are never together

$ \begin{aligned} & =\text { Total } - \text { Together } \\ \\ & =({ }^{10} C_3 \times{ }^{7} C_3+{ }^{10} C_4 \times{ }^{7} C_2)-({ }^{10} C_4 \times{ }^{5} C_0+{ }^{10} C_3 \times{ }^{5} C_1) \\ \\ & =(120 \times 35+210 \times 21)-(210+120 \times 5) \\ \\ & =4200+4410-(210+600) \\ \\ & =8610-810=7800 \end{aligned} $

Solution

Since, there are $2$ white, $3$ black and $4$ red balls. It is given that atleast one black ball is to be included in the draw.

$\therefore \ $ Required number of ways

$={ }^{3} C_1 \times{ }^{6} C_2+{ }^{3} C_2 \times{ }^{6} C_1+{ }^{3} C_3$

$ \begin{aligned} & =3 \times 15+3 \times 6+1 \\ \\ & =45+18+1=64 \end{aligned} $

Solution

False

Required number of lines $={ }^{12} C_2-{ }^{5} C_2+1$

Solution

False

Required number of ways $=5^{3}=125$

Solution

False

Arrangement of $n$ things, taken $r$ at a time in which $m$ things occur together, we considered these $m$ things as $1$ group.

$\therefore \ $ Number of object $=(r-m+1)$

Number of arrangements $=(r-m+1)!$

Now, combination of $n$ things taken $r$ at a time in which $m$ things occur $= {}^{n-m} C_{r-m}$

$\therefore \ $ Required number of arrangements $=(r-m+1) ! \times {}^{n-m} C_{r-m}$

Solution

True

There are three types of animals and stalls available for $12$ animals.

Number of ways of loading $=3^{12}$

Solution

True

If some or all objects taken at a time, then number of selection would be

$ { }^{n} C_1+{ }^{n} C_2+{ }^{n} C_3+\ldots+{ }^{n} C_{n}=2^{n}-1 \quad[\because{ }^{n} C_0+{ }^{n} C_1+{ }^{n} C_2+\ldots+{ }^{n} C_{n}=2^{n}] $

Solution

True

Total number of selection $=[(4+1)(5+1)-1]-5$

$ \begin{aligned} & =(5 \times 6-1)-5 \\ \\ & =(30-1)-5=24 \end{aligned} $

Solution

True

After seating $4$ on one side and $3$ on the other side, we have to select out of $11 ; 5$ on one side and $6$ on the other side.

Now, remaining selecting of one half side $={ }^{(18-4-3)} C_5={ }^{11} C_5$

and the other half side $={ }^{(11-5)} C_6={ }^{6} C_6$

Total arrangements $={ }^{11} C_5 \times 9 ! \times{ }^{6} C_6 \times 9 !$

$ \begin{aligned} & =\dfrac{11 !}{5 ! 6 !} \times 9 ! \times 1 \times 9 ! \\ \\ & =\dfrac{11 !}{5 ! 6 !} \times 9 ! \times 9 ! \end{aligned} $

Solution

False

He can attempt questions in following manner

$ \begin{array}{|c|c|c|c|c|} \hline \text{Group (A)} & 2 & 3 & 4 & 5 \\ \hline \text{Group (B)} & 5 & 4 & 3 & 2 \\ \hline \end{array} $

Number of ways of attempting $7$ questions

$ \begin{aligned} & ={ }^{6} C_2 \times{ }^{6} C_5+{ }^{6} C_3 \times{ }^{6} C_4+{ }^{6} C_4 \times{ }^{6} C_3+{ }^{6} C_5 \times{ }^{6} C_2 \\ \\ & =2({ }^{6} C_2 \times{ }^{6} C_5+{ }^{6} C_3 \times{ }^{6} C_4) \\ \\ & =2(15 \times 6+20 \times 15) \\ \\ & =2(90+300) \\ \\ & =2 \times 390=780 \end{aligned} $

Solution

False

We have to select $3$ scheduled caste candidate out of $5$ in ${ }^{5} C_3$ ways.

and we have to select $9$ other candidates out of $22$ in ${ }^{22} C_9$ ways.

$\therefore \ $ Total number of selections $={ }^{5} C_3 \times{ }^{22} C_9$

Solution

There are three books of Mathematics, $4$ of Physics and $5$ of English.

(i) One book of each subject $={ }^{3} C_1 \times{ }^{4} C_1 \times{ }^{5} C_1$

$ =3 \times 4 \times 5=60 $

(ii) Atleast one book of each subject $=(2^{3}-1) \times(2^{4}-1) \times(2^{5}-1)$

$ =7 \times 15 \times 31=3255 $

(iii) Atleast one book of English $=$ Selection based on following manner

$ \begin{array}{|c|c|c|c|c|c|} \hline \text{English book} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{Other} & 11 & 10 & 9 & 8 & 7 \\ \hline \end{array} $

$=(2^{5}-1) \times 2^{7}$

$=128 \times 31=3968$

Hence the correct matches are

(i) $\longleftrightarrow $ (b), (ii) $\longleftrightarrow $ (c), (iii) $\longleftrightarrow $ (a).

Solution

(i) Boys and girls alternate

Total arrangements $=(5 !)^{2}+(5 !)^{2}$

(ii) No two girls sit together $= 6! \times 5 !$

(iii) All the girls sit together $= 5! \times 6!$

(iv) All the girls are never together $=10 !-5 ! 6!$

Hence the correct matches are

(i) $\longleftrightarrow $ (c), (ii) $\longleftrightarrow $ (a), (iii) $\longleftrightarrow $ (a), (iv) $\longleftrightarrow $ (b).

Solution

(i) We have to select $2$ professors out of $10$ and $3$ lecturers out of $20={ }^{10} C_2 \times{ }^{20} C_3$

(ii) When a particular professor included $={ }^{10}{ }^{-1} C_1 \times{ }^{20} C_3={ }^{9} C_1 \times{ }^{20} C_3$

(iii) When a particular lecturer included $={ }^{10} C_2 \times{ }^{19} C_2$

(iv) When a particular lecturer excluded $={ }^{10} C_2 \times{ }^{19} C_3$

Hence the correct matches are

(i) $\longleftrightarrow $ (d), (ii) $\longleftrightarrow $ (c), (iii) $\longleftrightarrow $ (b), (iv) $\longleftrightarrow $ (a).

Solution

(i) Total numbers of 4 digit formed with digits $1, 2, 3, 4, 5, 6, 7$

$ =7 \times 6 \times 5 \times 4=840 $

(ii) When a number is divisible by $2$. At its unit place only even numbers occurs. Total numbers $=4 \times 5 \times 6 \times 3=360$

(iii) A number is divisible by $25$ if it ends with $25,50,75,00$.

$\therefore \ $ Total such numbers $=4 \times 5 \times 2 \times 1$

(iv) A number is divisible by $4$ , If its last two digit is divisible by $4$ .

$\therefore \ $ Total such numbers $=200$

Hence the correct matches are

(i) $\longleftrightarrow $ (a), (ii) $\longleftrightarrow $ (c), (iii) $\longleftrightarrow $ (d), (iv) $\longleftrightarrow $ (b).

Solution

(i) $4$ letters are used at a time $={ }^{6} P_4=\dfrac{6 !}{2 !}=6 \times 5 \times 4 \times 3=360$

(ii) All letters used at a time $=6 !=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$

(iii) All letters used but first is vowel $=2 \times 5 ! =2 \times 5 \times 4 \times 3 \times 2 \times 1=240$

Hence the correct matches are

(i) $\longleftrightarrow $ (c), (ii) $\longleftrightarrow $ (a), (iii) $\longleftrightarrow $ (b).