Answer :

It is known that, ${ }^{n} C_a={ }^{n} C_b \Rightarrow a=b$ or $n=a+b$

Therefore,

$ \begin{aligned} & { }^{n} C_8={ }^{n} C_2 \Rightarrow n=8+2=10 \\ \\ & \therefore \ \ \ { }^{n} C_2={ }^{10} C_2=\dfrac{10 !}{2 !(10-2) !}=\dfrac{10 !}{2 ! 8 !}=\dfrac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}=45 \end{aligned} $

Answer :

$ \begin{aligned} & \text{(i)} \quad \dfrac{{ }^{2 n} C_3}{{ }^{n} C_3}=\dfrac{12}{1} \\ \\ & \Rightarrow \dfrac{(2 n) !}{3 !(2 n-3) !} \times \dfrac{3 !(n-3) !}{n !}=\dfrac{12}{1} \\ \\ & \Rightarrow \dfrac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) !} \times \dfrac{(n-3) !}{n(n-1)(n-2)(n-3) !}=12 \\ \\ & \Rightarrow \dfrac{2(2 n-1)(2 n-2)}{(n-1)(n-2)}=12 \\ \\ & \Rightarrow \dfrac{4(2 n-1)(n-1)}{(n-1)(n-2)}=12 \\ \\ & \Rightarrow \dfrac{(2 n-1)}{(n-2)}=3 \\ \\ & \Rightarrow 2 n-1=3(n-2) \\ \\ & \Rightarrow 2 n-1=3 n-6 \\ \\ & \Rightarrow 3 n-2 n=-1+6 \\ \\ & \Rightarrow n=5 \end{aligned} $

$\text{(ii)} \quad \dfrac{{ }^{2 n} C_3}{{ }^{n} C_3}=\dfrac{11}{1}$

$\Rightarrow \dfrac{(2 n) !}{3 !(2 n-3) !} \times \dfrac{3 !(n-3) !}{n !}=11$

$\Rightarrow \dfrac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) !} \times \dfrac{(n-3) !}{n(n-1)(n-2)(n-3) !}=11$

$\Rightarrow \dfrac{2(2 n-1)(2 n-2)}{(n-1)(n-2)}=11$

$\Rightarrow \dfrac{4(2 n-1)(n-1)}{(n-1)(n-2)}=11$

$\Rightarrow \dfrac{4(2 n-1)}{n-2}=11$

$\Rightarrow 4(2 n-1)=11(n-2)$

$\Rightarrow 8 n-4=11n -22$

$\Rightarrow 11n -8 n=-4+22$

$\Rightarrow 3 n=18$

$\Rightarrow n=6$

Answer :

For drawing one chord on a circle, only $2$ points are required.

To know the number of chords that can be drawn through the given $21$ points on a circle, the number of combinations have to be counted.

Therefore, there will be as many chords as there are combinations of $21$ points taken $2$ at a time.

Thus, required number of chords $={ }^{21} C_2=\dfrac{21 !}{2 !(21-2) !}=\dfrac{21 !}{2 ! 19 !}=\dfrac{21 \times 20}{2}=210$

Answer :

A team of $3$ boys and $3$ girls is to be selected from $5$ boys and $4$ girls.

$3$ boys can be selected from $5$ boys in ${ }^{5} C_3$ ways.

$3$ girls can be selected from $4$ girls in ${ }^{4} C_3$ ways.

Therefore, by multiplication principle, number of ways in which a team of $3$ boys and $3$ girls can be selected

$ ={ }^{5} C_3 \times{ }^{4} C_3=\dfrac{5 !}{3 ! 2 !} \times \dfrac{4 !}{3 ! 1 !} $

$=\dfrac{5 \times 4 \times 3 !}{3 ! \times 2} \times \dfrac{4 \times 3 !}{3 !}$

$=10 \times 4=40$

Answer :

There are a total of $6$ red balls, $5$ white balls, and $5$ blue balls.

$9$ balls have to be selected in such a way that each selection consists of $3$ balls of each colour.

Here,

$3$ balls can be selected from $6$ red balls in ${ }^{6} C_3$ ways.

$3$ balls can be selected from $5$ white balls in ${ }^{5} C_3$ ways.

$3$ balls can be selected from $5$ blue balls in ${ }^{5} C_3$ ways.

Thus, by multiplication principle, required number of ways of selecting $9$ balls

$ \begin{aligned} & ={ }^{6} C_3 \times{ }^{5} C_3 \times{ }^{5} C_3=\dfrac{6 !}{3 ! 3 !} \times \dfrac{5 !}{3 ! 2 !} \times \dfrac{5 !}{3 ! 2 !} \\ \\ & =\dfrac{6 \times 5 \times 4 \times 3 !}{3 ! \times 3 \times 2} \times \dfrac{5 \times 4 \times 3 !}{3 ! \times 2 \times 1} \times \dfrac{5 \times 4 \times 3 !}{3 ! \times 2 \times 1} \\ \\ & =20 \times 10 \times 10=2000 \end{aligned} $

Answer :

In a deck of $52$ cards, there are $4$ aces. A combination of $5$ cards have to be made in which there is exactly one ace.

Then, one ace can be selected in ${ }^{4} C_1$ ways and the remaining $4$ cards can be selected out of the $48$ cards in ${ }^{48} C_4$ ways.

$ \begin{aligned} & ={ }^{48} C_4 \times{ }^{4} C_1=\dfrac{48 !}{4 ! 44 !} \times \dfrac{4 !}{1 ! 3 !} \\ \\ & =\dfrac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} \times 4 \end{aligned} $

Thus, by multiplication principle, required number of $5$ card combinations $=778320$

Answer :

Out of $17$ players, $5$ players are bowlers.

A cricket team of $11$ players is to be selected in such a way that there are exactly $4$ bowlers.

$4$ bowlers can be selected in ${ }^{5} C_4$ ways and the remaining $7$ players can be selected out of the $12$ players in ${ }^{12} C_7$ ways.

Thus, by multiplication principle, required number of ways of selecting cricket team

$ ={ }^{5} C_4 \times{ }^{12} C_7=\dfrac{5 !}{4 ! 1 !} \times \dfrac{12 !}{7 ! 5 !}=5 \times \dfrac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}=3960 $

Answer :

There are $5$ black and $6$ red balls in the bag.

$2$ black balls can be selected out of $5$ black balls in ${ }^{5} C_2$ ways and $3$ red balls can be selected out of $6$ red balls in ${ }^{6} C_3$ ways.

Thus, by multiplication principle, required number of ways of selecting $2$ black and $3$ red

balls $={ }^{5} C_2 \times{ }^{6} C_3=\dfrac{5 !}{2 ! 3 !} \times \dfrac{6 !}{3 ! 3 !}=\dfrac{5 \times 4}{2} \times \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1}=10 \times 20=200$

Answer :

There are $9$ courses available out of which, $2$ specific courses are compulsory for every student.

Therefore, every student has to choose $3$ courses out of the remaining $7$ courses. This can be chosen in ${ }^{7} C_3$ ways. Thus, required number of ways of choosing the programme

$ ={ }^{7} C_3=\dfrac{7 !}{3 ! 4 !}=\dfrac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}=35 $