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A group consists of 3 couples in which each of the 3 men have one wife each.In how many ways could they arranged in a straight line so that the men and women occupy alternate position ?
- a)216
- b)125
- c)256
- d)72
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
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A group consists of 3 couples in which each of the 3 men have one wife...
Answer – D.72 Explanation : 3!*3! + 3!*3! = 36+36 = 72
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A group consists of 3 couples in which each of the 3 men have one wife...
Problem: A group consists of 3 couples in which each of the 3 men have one wife each. In how many ways could they arranged in a straight line so that the men and women occupy alternate position?
Solution:
We have 3 couples, so there are 3 men and 3 women. To form a straight line, the men and women have to alternate positions. Let us first arrange the 3 men in alternate positions which can be done in 3! ways. Now, we have to arrange the women in alternate positions.
Case 1: First woman is placed in any of the three alternate positions of the first man.
This can be done in 3 ways. And the second woman can be placed in any of the two remaining alternate positions of the second man. This can be done in 2 ways. The third woman can be placed in any of the remaining alternate positions of the third man. This can be done in 1 way.
Therefore, the total number of ways of arranging the women in this case is 3 x 2 x 1 = 6.
Case 2: First woman is placed in any of the two remaining alternate positions of the second man.
This can be done in 2 ways. And the second woman can be placed in any of the remaining alternate positions of the third man. This can be done in 1 way. The third woman can be placed in the remaining alternate position of the first man.
Therefore, the total number of ways of arranging the women in this case is 2 x 1 x 1 = 2.
So, the total number of ways of arranging the men and women alternately is 3! x (6 + 2) = 72.
Hence, the correct answer is option D.
Solution:
We have 3 couples, so there are 3 men and 3 women. To form a straight line, the men and women have to alternate positions. Let us first arrange the 3 men in alternate positions which can be done in 3! ways. Now, we have to arrange the women in alternate positions.
Case 1: First woman is placed in any of the three alternate positions of the first man.
This can be done in 3 ways. And the second woman can be placed in any of the two remaining alternate positions of the second man. This can be done in 2 ways. The third woman can be placed in any of the remaining alternate positions of the third man. This can be done in 1 way.
Therefore, the total number of ways of arranging the women in this case is 3 x 2 x 1 = 6.
Case 2: First woman is placed in any of the two remaining alternate positions of the second man.
This can be done in 2 ways. And the second woman can be placed in any of the remaining alternate positions of the third man. This can be done in 1 way. The third woman can be placed in the remaining alternate position of the first man.
Therefore, the total number of ways of arranging the women in this case is 2 x 1 x 1 = 2.
So, the total number of ways of arranging the men and women alternately is 3! x (6 + 2) = 72.
Hence, the correct answer is option D.
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