Question
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Answer: Option D
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
∴ the Required number of ways
$$\eqalign{
& = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + \left( {{}^7{C_4} \times {}^6{C_1}} \right) + \left( {{}^7{C_5}} \right) \cr
& = \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \frac{{6 \times 5}}{{2 \times 1}}} \right) + \left( {{}^7{C_3} \times {}^6{C_1}} \right) + \left( {{}^7{C_2}} \right) \cr
& = 525 + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right) + \left( {\frac{{7 \times 6}}{{2 \times 1}}} \right) \cr
& = \left( {525 + 210 + 21} \right) \cr
& = 756 \cr} $$
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We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
∴ the Required number of ways
$$\eqalign{
& = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + \left( {{}^7{C_4} \times {}^6{C_1}} \right) + \left( {{}^7{C_5}} \right) \cr
& = \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \frac{{6 \times 5}}{{2 \times 1}}} \right) + \left( {{}^7{C_3} \times {}^6{C_1}} \right) + \left( {{}^7{C_2}} \right) \cr
& = 525 + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right) + \left( {\frac{{7 \times 6}}{{2 \times 1}}} \right) \cr
& = \left( {525 + 210 + 21} \right) \cr
& = 756 \cr} $$
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