In how many different ways can 4 ladies and 4 gentlemen be

Treat the 4 ladies as one object, now you have 5 objects to arrange around a table (m1,m2,m3,m4,women). This can be done in (5-1)! ways
And there are 4! ways to arrange ladies among themselves

Answer = (4!)^2 = 576 or C

For concept's sake, if we were to do this the opposite way, how would we do it? Say we have (8-1)! of arranging without any conditions. Then it should be 7! - number of ways 2 women can sit together - number of ways three can sit together.

so for number of ways two can sit together I get: (4-1)! and then 4C3 (in how many ways can we place 3 women in 4 slots, since I tied two together * 2)

Number of ways 3 can sit together= seat the men in (4-1)! ways. * 4C2 (in how many ways can two women be placed in 4 slots, since I tied three women together this time)*3! (for the number of arrangements of three women ties together)

This doesn't give me the correct answer. Where have I gone wrong?

_ _ _ _ _ _ _ _ <---- 8 spots

Combination:
Since the four women must be together there is 4C4 ways we can choose seats for them.

Permutation:
Among the women, there are 4! ways we can arrange them.
Likewise among men there are 4! ways that they can be arranged

4! x 4! = 576.

Answer is C.

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