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?
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?
To solve this problem, we can use combinations to determine the number of ways to select the committee members with the given conditions.
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Selecting 3 Men and 2 Women:
- If at least 3 men are to be on the committee, we can have either 3 men and 2 women or 4 men and 1 woman on the committee.
- The number of ways to choose 3 men out of 7 is \binom{7}{3}.
- The number of ways to choose 2 women out of 6 is \binom{6}{2}.
- Therefore, the total number of ways to select 3 men and 2 women is \binom{7}{3} \times \binom{6}{2}.
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Selecting 4 Men and 1 Woman:
- The number of ways to choose 4 men out of 7 is \binom{7}{4}.
- The number of ways to choose 1 woman out of 6 is \binom{6}{1}.
- Therefore, the total number of ways to select 4 men and 1 woman is \binom{7}{4} \times \binom{6}{1}.
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Total Number of Ways:
- Add the two cases together to find the total number of ways to form the committee: \binom{7}{3} \times \binom{6}{2} + \binom{7}{4} \times \binom{6}{1}.
Now, let’s calculate the total number of ways using the combinations formula:
\binom{7}{3} = \frac{7!}{3!(7-3)!} = 35
\binom{6}{2} = \frac{6!}{2!(6-2)!} = 15
\binom{7}{4} = \frac{7!}{4!(7-4)!} = 35
\binom{6}{1} = \frac{6!}{1!(6-1)!} = 6
Substitute these values into the total number of ways equation:
35 \times 15 + 35 \times 6 = 525 + 210 = 735
Therefore, there are 735 ways to form the committee with at least 3 men.