Answer :

In the word DAUGHTER, there are $ 3 $ vowels namely, $A, \ U,$ and $E,$ and $5 $ consonants namely, $D,$ $G,$ $H,$ $T,$ and $R.$

Number of ways of selecting $ 2 $ vowels out of $ 3 $ vowels $={ }^{3} C_2=3$

Number of ways of selecting $ 3 $ consonants out of $ 5 $ consonants $={ }^{5} C_3=10$

Therefore, number of combinations of $ 2 $ vowels and $ 3 $ consonants $=3 \times 10=30$

Each of these $ 30 $ combinations of $ 2 $ vowels and $ 3 $ consonants can be arranged among themselves in $5 !$ ways.

Hence, required number of different words $=30 \times 5 !=3600$

Answer :

In the word EQUATION, there are $ 5 $ vowels, namely, $A, E, I, O$, and $U$, and $ 3 $ consonants, namely, $ Q,$ $T$, and $N$. Since all the vowels and consonants have to occur together, both $(AEIOU)$ and $(QTN)$ can be assumed as single objects.

Then, the permutations of these $ 2 $ objects taken all at a time are counted. This number would be ${ }^{2} P_2=2 !$ Corresponding to each of these permutations, there are $5 !$ permutations of the five vowels taken all at a time and $3 !$ permutations of the $ 3 $ consonants taken all at a time.

Hence, by multiplication principle, required number of words $=2 ! \times 5 ! \times 3 !$ $=1440$

Answer :

A committee of $ 7 $ has to be formed from $ 9 $ boys and $ 4 $ girls.

(i). Since exactly $ 3 $ girls are to be there in every committee, each committee must consist of $(7-3)$ boys only.

Thus, in this case, required number of ways $={ }^{4} C_3 \times{ }^{9} C_4=\dfrac{4 !}{3 ! 1 !} \times \dfrac{9 !}{4 ! 5 !}=4 \times \dfrac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !}=504$

(ii) Since at least $ 3 $ girls are to be there in every committee, the committee can consist of

Answer :

In the given word EXAMINATION, there are $ 11 $ letters out of which, $A, I$, and $N$ appear $ 2 $ times and all the other letters appear only once.

The words that will be listed before the words starting with $E$ in a dictionary will be the words that start with $A$ only.

Therefore, to get the number of words starting with $ A,$ the letter A is fixed at the extreme left position, and then the remaining $ 10 $ letters taken all at a time are rearranged.

Since there are $ 2 I$ and $2 N$ in the remaining $ 10 $ letters,

Number of words starting with $A=\dfrac{10 !}{2 ! 2 !}=907200$

Thus, the required numbers of words is $907200 .$

Answer :

A number is divisible by $ 10 $ if its units digits is $ 0 $ .

Therefore, $ 0 $ is fixed at the units place.

Therefore, there will be as many ways as there are ways of filling $ 5 $ vacant places

$\begin{array}{|c|c|c|c|c|c|} \hline & & & & & 0 \\ \hline \end{array}$

by the remaining 5 digits (i.e., $ 1,$ $ 3,$ $ 5,$ $7$ and $9$ ).

The 5 vacant places can be filled in $5 !$ ways.

Hence, required number of 6-digit numbers $=5 !=120$

Answer :

2 different vowels and $ 2 $ different consonants are to be selected from the English alphabet.

Since there are $ 5 $ vowels in the English alphabet, number of ways of selecting $ 2 $ different vowels from the alphabet

$ ={ }^{5} C_2=\dfrac{5 !}{2 ! 3 !}=10 $

Since there are $ 21 $ consonants in the English alphabet, number of ways of selecting $ 2 $ different consonants from the alphabet

$ ={ }^{21} C_2=\dfrac{21 !}{2 ! 19 !}=210 $

Therefore, number of combinations of $ 2 $ different vowels and $ 2 $ different consonants $=10 \times 210=2100$

Each of these $ 2100 $ combinations has $ 4 $ letters, which can be arranged among themselves in $4!$ ways.

Therefore, required number of words $=2100 \times 4 !=50400$

Answer :

It is given that the question paper consists of $ 12 $ questions divided into two parts Part I and Part II, containing $ 5 $ and $ 7 $ questions, respectively.

A student has to attempt $ 8 $ questions, selecting at least $ 3 $ from each part.

This can be done as follows.

Answer :

From a deck of $ 52 $ cards, $ 5 $ card combinations have to be made in such a way that in each selection of $ 5 $ cards, there is exactly one king.

In a deck of $ 52 $ cards, there are $ 4 $ kings.

$1$ king can be selected out of $ 4 $ kings in ${ }^{4} C_1$ ways.

$4$ cards out of the remaining $ 48 $ cards can be selected in ${ }^{48} C_4$ ways.

Thus, the required number of $5$ card combinations is ${ }^{4} C_1 \times{ }^{48} C_4$.

Answer :

total number of people 9.

women occupy only even places women can occupy only four even places

$ \begin{aligned} & \mathrm{M} \underline{W} \mathrm{M} \underline{W} \mathrm{M} \underline{W} \mathrm{M} \underline{W} \mathrm{M} \\ \\ & 4 \text { women can sit in } 4 \text { positions } \\ \\ & \therefore \quad \text { total no. of ways }={}^4P_4 = \frac{4!}{(4-4)!} =24 \text { ways } \\ \\ & 5 \text { men can sit in } 5 \text { positions } \\ \\ & \therefore \quad \text { total no. of ways }= {}^5P_5 =\frac{5!}{(5-5)!} \\ \\ & \therefore \quad \text { total no. of agreements }=24 \times 120 =2880 \text { ways } \end{aligned} $

Answer :

From the class of $ 25 $ students, $ 10 $ are to be chosen for an excursion party.

Since there are $ 3 $ students who decide that either all of them will join or none of them will join, there are two cases.

Case I: All the three students join.

Then, the remaining $ 7 $ students can be chosen from the remaining $ 22 $ students in ${ }^{22} C_7$ ways.

Case II: None of the three students join.

Then, $ 10 $ students can be chosen from the remaining $ 22 $ students in ${ }^{22} C _{10}$ ways.

Thus, required number of ways of choosing the excursion party is ${ }^{22} C_7+{ }^{22} C _{10}$.

Answer :

In the given word ASSASSINATION, the letter A appears $ 3 $ times, $S$ appears $ 4 $ times, $I $ appears $ 2 $ times, $N$ appears $ 2 $ times, and all the other letters appear only once.

Since all the words have to be arranged in such a way that all the $S$ are together, $SSSS$ is treated as a single object for the time being. This single object together with the remaining $ 9 $ objects will account for $ 10 $ objects.

These $ 10 $ objects in which there are $ 3 A$ , $ 2I$ , and $ 2N$ can be arranged in $\dfrac{10 !}{3 ! 2 ! 2 !}$ ways.

Thus, required number of ways of arranging the letters of the given word $=\dfrac{10 !}{3 ! 2 ! 2 !}=151200$