Question

A students council of 6 members is to be formed from a selection pool of 8 males and 9 females . How many committee's will have at least 4 females?​

Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Given that

A students council of 6 members is to be formed from a selection pool of 8 males and 9 females.

We have to choose 6 members so that there are atleast 4 females.

So, following case arises

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Male (8)& \bf Female(9) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 & \sf 4 \\ \\ \sf 1 & \sf 5 \\ \\ \sf 0 & \sf 6 \end{array}} \\ \end{gathered}$

We know,

The number of ways in which r objects can be selected from n distinct objects are

$\boxed{\tt{ \: ^nC_r \: = \: \frac{n!}{r! \: (n - r)! \: }}}$

So, number of ways of selecting 2 males and 4 females from 8 males and 9 females is

$\rm \:  =  \: ^8C_2 \: \times \: ^9C_4$

$\rm \:  =  \: \dfrac{8 \times 7}{2} \times \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$

$\rm \:  =  \: 28 \times 126$

$\rm \:  =  \: 3328$

Now, number of ways of selecting 1 males and 5 females from 8 males and 9 females is

$\rm \:  =  \: ^8C_1 \: \times \: ^9C_5$

$\rm \:  =  \: \dfrac{8}{1} \times \dfrac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1}$

$\rm \:  =  \: 8 \times 126$

$\rm \:  =  \: 1008$

Again, number of ways of selecting 0 males and 6 females from 8 males and 9 females is

$\rm \:  =  \: ^8C_0 \: \times \: ^9C_6$

$\rm \:  =  \: 1 \: \times \: ^9C_3$

$\rm \:  =  \: \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}$

$\rm \:  =  \: 84$

Hence,

Total number of ways in which a students council of 6 members is to be formed from a selection pool of 8 males and 9 females, comprised of atleast 4 females is

$\rm \:  =  \: 3328 + 1008 + 84$

$\rm \:  =  \: 4420$

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Additional Information :-

$\boxed{\tt{ \: ^{n}C_{r} \: = \: ^{n}C_{n - r} \: }}$

$\boxed{\tt{ \: ^{n}C_{0} \: = \: ^{n}C_{n} \: = \: 1 \: }}$

$\boxed{\tt{ \: ^{n}C_{1} \: = \: ^{n}C_{n - 1} \: = \: n \: }}$

$\boxed{\tt{ \: ^{n}C_{r} \: + \: ^{n}C_{r - 1} \: = \: ^{n + 1}C_{r} \: }}$

$\boxed{\tt{ \frac{ \: ^{n}C_{r} \: }{ \: ^{n}C_{r - 1} \: } = \: \frac{n - r + 1}{r} \: }}$