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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?
- a)564
- b)645
- c)735
- d)756
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
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To form a committee, we need to select 5 people from a group of 7 men and 6 women. The condition is that there should be at least 3 men on the committee.
Let's consider the different possibilities:
1. Selecting 3 men and 2 women:
- We can select 3 men from the 7 available men in 7C3 ways (7 choose 3).
- We can select 2 women from the 6 available women in 6C2 ways (6 choose 2).
- The total number of ways to select 3 men and 2 women is 7C3 * 6C2.
2. Selecting 4 men and 1 woman:
- We can select 4 men from the 7 available men in 7C4 ways.
- We can select 1 woman from the 6 available women in 6C1 ways.
- The total number of ways to select 4 men and 1 woman is 7C4 * 6C1.
3. Selecting all 5 men:
- We can select all 5 men from the 7 available men in 7C5 ways.
Now, let's calculate the total number of ways to form the committee by adding up the possibilities:
Total number of ways = 7C3 * 6C2 + 7C4 * 6C1 + 7C5
Calculating the combinations:
7C3 = 7! / (3! * (7-3)!) = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
6C2 = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
7C4 = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
6C1 = 6! / (1! * (6-1)!) = 6! / (1! * 5!) = 6 / 1 = 6
7C5 = 7! / (5! * (7-5)!) = 7! / (5! * 2!) = (7 * 6) / (2 * 1) = 21
Substituting the values:
Total number of ways = 35 * 15 + 35 * 6 + 21 = 525 + 210 + 21 = 756
Therefore, there are 756 ways to form the committee with at least 3 men. Hence, the correct answer is option D) 756.
Let's consider the different possibilities:
1. Selecting 3 men and 2 women:
- We can select 3 men from the 7 available men in 7C3 ways (7 choose 3).
- We can select 2 women from the 6 available women in 6C2 ways (6 choose 2).
- The total number of ways to select 3 men and 2 women is 7C3 * 6C2.
2. Selecting 4 men and 1 woman:
- We can select 4 men from the 7 available men in 7C4 ways.
- We can select 1 woman from the 6 available women in 6C1 ways.
- The total number of ways to select 4 men and 1 woman is 7C4 * 6C1.
3. Selecting all 5 men:
- We can select all 5 men from the 7 available men in 7C5 ways.
Now, let's calculate the total number of ways to form the committee by adding up the possibilities:
Total number of ways = 7C3 * 6C2 + 7C4 * 6C1 + 7C5
Calculating the combinations:
7C3 = 7! / (3! * (7-3)!) = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
6C2 = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
7C4 = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
6C1 = 6! / (1! * (6-1)!) = 6! / (1! * 5!) = 6 / 1 = 6
7C5 = 7! / (5! * (7-5)!) = 7! / (5! * 2!) = (7 * 6) / (2 * 1) = 21
Substituting the values:
Total number of ways = 35 * 15 + 35 * 6 + 21 = 525 + 210 + 21 = 756
Therefore, there are 756 ways to form the committee with at least 3 men. Hence, the correct answer is option D) 756.
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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?a)564b)645c)735d)756e)None of theseCorrect answer is option 'D'. Can you explain this answer?
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