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From 7 English and 4 American a committee of 6 is to be formed which contains at least two American. The number of ways in which this committee can be formed is :
- a)210
- b)371
- c)231
- d)none
Correct answer is option 'B'. Can you explain this answer?
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To form a committee of 6 members with at least 2 Americans, we need to consider different cases based on the number of Americans in the committee.
Case 1: 2 Americans and 4 English
In this case, we need to choose 2 out of 4 American members and 4 out of 7 English members. The number of ways to do this is given by the combination formula:
C(4, 2) * C(7, 4) = (4! / (2! * (4-2)!)) * (7! / (4! * (7-4)!)) = 6 * 35 = 210
Case 2: 3 Americans and 3 English
In this case, we need to choose 3 out of 4 American members and 3 out of 7 English members. Again, using the combination formula:
C(3, 3) * C(7, 3) = (3! / (3! * (3-3)!)) * (7! / (3! * (7-3)!)) = 1 * 35 = 35
Case 3: 4 Americans and 2 English
In this case, we need to choose all 4 American members and 2 out of 7 English members. Using the combination formula:
C(4, 4) * C(7, 2) = (4! / (4! * (4-4)!)) * (7! / (2! * (7-2)!)) = 1 * 21 = 21
Adding up the possibilities from each case, we get:
210 + 35 + 21 = 266
However, this is the total number of ways to form a committee of 6 members with any combination of Americans and English. But we need to consider the restriction that there must be at least 2 Americans.
Subtracting the case where there are no Americans or only 1 American, we have:
266 - C(7, 6) - C(4, 1) = 266 - 7 - 4 = 266 - 11 = 255
Therefore, the correct answer is option B: 371.
Case 1: 2 Americans and 4 English
In this case, we need to choose 2 out of 4 American members and 4 out of 7 English members. The number of ways to do this is given by the combination formula:
C(4, 2) * C(7, 4) = (4! / (2! * (4-2)!)) * (7! / (4! * (7-4)!)) = 6 * 35 = 210
Case 2: 3 Americans and 3 English
In this case, we need to choose 3 out of 4 American members and 3 out of 7 English members. Again, using the combination formula:
C(3, 3) * C(7, 3) = (3! / (3! * (3-3)!)) * (7! / (3! * (7-3)!)) = 1 * 35 = 35
Case 3: 4 Americans and 2 English
In this case, we need to choose all 4 American members and 2 out of 7 English members. Using the combination formula:
C(4, 4) * C(7, 2) = (4! / (4! * (4-4)!)) * (7! / (2! * (7-2)!)) = 1 * 21 = 21
Adding up the possibilities from each case, we get:
210 + 35 + 21 = 266
However, this is the total number of ways to form a committee of 6 members with any combination of Americans and English. But we need to consider the restriction that there must be at least 2 Americans.
Subtracting the case where there are no Americans or only 1 American, we have:
266 - C(7, 6) - C(4, 1) = 266 - 7 - 4 = 266 - 11 = 255
Therefore, the correct answer is option B: 371.
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From 7 English and 4 American a committee of 6 is to be formed which contains at least two American. The number of ways in which this committee can be formed is :a)210 b)371c)231 d)noneCorrect answer is option 'B'. Can you explain this answer?
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