Of 4 women and 3 men in a table row, how many ways can the chair be arranged such that two men cannot seat together?

Answer 1

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alternate woman man, with women at each end of the table row

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9 different ways, either keep the women all apart or have the 2 women who are together at the beginning, the end, or 4 different ways with 2 women together where they are not at the end, or have males at both ends with 3 women together inside, in 2 ways

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Answer 2

The answer is 144. Here is one way to go about solving this problem:

1. Let's say that the first seat can be occupied by any of the 4 women: so, 4 different persons can sit there (4)
2. The second seat can be occupied by any of the 3 men: so, 3 different persons can sit there (3)
3. The third seat can be occupied by any of the 3 women left (remember, one of them is already sitting on the first seat :) (3)
4. The fourth seat can be occupied by any of the two men left (2)
5. The fifth seat can be occupied by any of the two women left (2)
6. The sixth seat can be occupied by the only man left (1)
7. The seventh seat can be occupied by the only woman left (1)

Now, we multiply the number of people that can seat on each of the seven seats to get all the possible combinations: 4 x 3 x 3 x 2 x 2 x 1 x 1 = 144

You can start with any of the men in the first seat instead of any of the women, but in the end, you will still have 144 possible combinations.

This problem deals with the topic of Combinations, and solving it can be simplified by multiplying 4! by 3!, where 4! represents all the possible combinations that women can sit, while 3! represents all the possible combinations that men can seat.