Question

Answer the question given in attachment ​

Answer 1

EXPLANATION.

A committee of 7 is to be formed from 9 bòys and 4 gìrls.

In how many ways this is be done when the committee consists of.

(1) = Exactly 3 gìrls.

(2) = At least 3 gìrls.

As we know that,

(1) = Exactly 3 gìrls.

Total number = 7.

Number of gìrls = 4.

Number of choose = 3.

Total number of ways = ⁴C₃ x ⁹C₄.

⇒ ⁴C₃ x ⁹C₄.

⇒ 4!/(3!)(4 - 3)! x 9!/(4)!(9 - 4)!.

⇒ 4!/3! x 9!/(4!)(5!).

⇒ 4 x 3!/3! x (9 x 8 x 7 x 6 x 5!)/(4 x 3 x 2 x 1)(5!).

⇒ 4 x (9 x 8 x 7 x 6)/(4 x 3 x 2).

⇒ 4 x (9 x 8 x 7 x 6)/24.

⇒ 9 x 8 x 7 = 504.

(2) = At least 3 gìrls.

First possibility is,

Total number = 7.

Number of gìrls = 4.

Number of choose = 3.

Total number of ways = ⁴C₃ x ⁹C₄.

⇒ ⁴C₃ x ⁹C₄.

⇒ 4!/(3!)(4 - 3)! x 9!/(4)!(9 - 4)!.

⇒ 4!/3! x 9!/(4!)(5!).

⇒ 4 x 3!/3! x (9 x 8 x 7 x 6 x 5!)/(4 x 3 x 2 x 1)(5!).

⇒ 4 x (9 x 8 x 7 x 6)/(4 x 3 x 2).

⇒ 4 x (9 x 8 x 7 x 6)/24.

⇒ 9 x 8 x 7 = 504.

Second possibility is.

We can select 4 gìrls and 3 bòys.

⇒ ⁴C₄ x ⁹C₃.

⇒ 4!/(4!)(4 - 4)! x 9!/(3!)(9 - 3)!.

⇒ 4!/(4!)(0!) x 9!/3! x 6!.

⇒ 1 x (9 x 8 x 7 x 6!)?(3 x 2 x 1)(6!).

⇒ 1 x 9 x 8 x 7/6.

⇒ 84.

Total number of ways = 504 + 84 = 588 ways.

Answer 2

$\mathbb{\colorbox{red}{\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox{pink}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox{peach}{Answer:-}}}}}}}}}}}}}}}$

(i) when committee consists of exactly 3 girls. Then remaining will be 4 boys in the committee.

$So, the \: required \: no.of \: ways \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = {}^{9} C4 \times {}^{4} C3$

$= \frac{9 }{4! \:5! } \times \frac{4!}{3! \: 1!} = \frac{9 \times 8 \times 7 \times 6}{6} = 504$

(ii) when committee consists of atleast 3 girls

i.e., it may be 3 girls or 4 girls. Then there are two cases. One when there are 3 girls in a committee, then the boys will be 4 & Second when there are 4 girls in a committee, then the boys will be 3.

$So, No. of \: ways \: in \: first \: case \\ \: \: = {}^{4} C3 \times {}^{9} C4,$

$No. of \: ways \: in \: second \: case \\ \: \: \: \: \: \: \: \: = {}^{4} C4 \times {}^{9} C3</p><p>$

$∴ Required \: no. of \: ways \\ \: \: \: \: \: = {}^{4} C3 \times {}^{9} C4 \: + {}^{4} C4 \times {}^{9} C3$

$= \frac{4!}{3! \: 1!} \times \frac{9!}{4! \: 5!} + \frac{4!}{4! \: 0!} \times \frac{9!}{6! \: 3!}$

$= \frac{9 \times 8 \times 7 \times 6}{6} + \frac{9 \times 8 \times 7}{6}$

$\huge\red{A}\pink{N}\orange{S}\green{W}\blue{E}\gray{R:-}$

$\huge\boxed{\fcolorbox{blue}{blue}{504+84=588}}$