A committee of 10 members has to be formed from 15 men and 7 women. In how
many ways can this be done when the committee cons

2 thoughts on “A committee of 10 members has to be formed from 15 men and 7 women. In how<br />many ways can this be done when the committee cons”

1. required Answer :-

   ☯ GIVEN ,

   A committee of 10 members has to be formed from 15 men and 7 women.

   ✿ Solution ,

   [tex] = \frac{7!}{6!} \times \frac{15!}{4! \times 11!} + \frac{7!}{7!} \times \frac{15!}{3! \times 21!} [/tex]

   [tex] = 7 \times \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2} + 1 \times \frac{15 \times 14 \times 13}{3 \times 2} [/tex]

   ⟹[tex]now \: 7 \times 1365 + 455[/tex]

   ⟹[tex]9555 + 445[/tex]

   ⟹[tex]10010[/tex]

   Hence , 10010 is the Answer 🙂

   Reply

2. Answer:

   Number of ways = 10010

   Step-by-step explanation:

   Given:

   • A committee of 10 members is to be formed from 15 men and 7 women.

   To Find:

   • Number of ways the committee can be formed if it consists of atleast 6 women

   Solution:

   By given the committee must not exceed 10 members and should contain atleast 6 women.

   Thus the committee can be selected in the following ways:

   6 women and 4 men or 7 women and 3 men

   This can be represented as,

   [tex]\tt Total\:number\:of\:ways=\:^7C_6\times \:^{15}C_4+\: ^7C_7+\: ^{15}C_3[/tex]

   We know that,

   [tex]\boxed{\tt ^nC_r=\dfrac{n!}{r!(n-r)!} }[/tex]

   Therefore we get,

   [tex]\tt Total\:number\:of\:ways=\dfrac{7!}{6!}\times \dfrac{15!}{4!\times 11!} +\dfrac{7!}{7!} \times \dfrac{15!}{3!\times 12!}[/tex]

   [tex]\tt \implies 7\times \dfrac{15\times 14\times 13\times 12}{4\times 3\times 2} +1\times \dfrac{15\times 14\times 13}{3\times 2}[/tex]

   [tex]\tt \implies 7\times1365+ 455[/tex]

   [tex]\tt \implies 9555+455[/tex]

   [tex]\tt \implies 10010[/tex]

   Hence the committee can be formed in 10010 ways.

   Reply