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Standard XII

Mathematics

Combination

There are 7...

Question

There are $7$ men and $5$ women who are eligible for being included in a committee of $4$ men and $3$ women. In how many ways can the committee be formed if one women $W_{1}$ refuses to be in the committee if another particular women $W_{2}$ be included?

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Solution

Total there are $7$ men and $5$ women

Committee of $4$ men and $3$ women is to be formed

Men can be selected in ${7_{C}}_{4}$ $= 35$

$W 1$ refuses and including $W 2$ number of ways for women are ${4_{C}}_{1}$ $= 4$

Total ways $= 35$ $\times$ $4 = 280$

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There are 7 men and 5 women who are eligible for being included in a committee of 4 men and 3 women. In how many ways can the committee be formed if one women W1 refuses to be in the committee if another particular women W2 be included?

Total there are 7 men and 5 womenCommittee of 4 men and 3 women is to be formedMen can be selected in 7C4 =35W1 refuses and including W2 number of ways for women are 4C1 =4Total ways =35 × 4=280

There are $7$ men and $5$ women who are eligible for being included in a committee of $4$ men and $3$ women. In how many ways can the committee be formed if one women $W_{1}$ refuses to be in the committee if another particular women $W_{2}$ be included?

Total there are $7$ men and $5$ women

Committee of $4$ men and $3$ women is to be formed

Men can be selected in ${7_{C}}_{4}$ $= 35$

$W 1$ refuses and including $W 2$ number of ways for women are ${4_{C}}_{1}$ $= 4$

Total ways $= 35$ $\times$ $4 = 280$