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Find the number of ways this can be arranged in which no 2 women and no 2 men sit together given 4 men and 3 women are seated in a dinner table?

@Edit: They are seating in a row dinner table

I have applied 2 approaches in this question and found both approaches differ according to the Ans The Ans is 144

I want to ask why the second approach is wrong?

My 1st Approach:

When i do arrangement like this: first arrange all 3 women at alternate places i.e at 2 4 6 or 1,3,5.

Now the next 4 places ca be filled in 4! ways.

So,the required arrangement=4! .3!=144 Ans

2 Approach

When i do arrangement like this: first arrange all 4 men at alternate places i.e at 2 4 6 8 or 1 3 5 7.

Now,arrange the 3 Women can be placed at 5 places=5P3=60

The required arrangement=60 . 24=1440

Gerry Myerson
  • 179,216
Jack
  • 752

2 Answers2

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You don't have place number $8$ as you stated in 2nd approach. Actually this is easy because $7$-seated table forces us to put all men in spots $1,3,5,7$ and all women in spots $2,4,6$, so the question becomes:

In how many ways can I rotate 4 men and 3 women (in groups)?

Is is $4! \cdot 3! = 144$.

Wojciech Karwacki
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There are only 3 women, the only way one could possibly satisfy the conditions of the question is that the ends are occupied by men. Supposing you began with a women at the first position you ll get an arrangement like the following W M W M W _ _ Given that there are no more women left, the only option is to seat two men in the last two positions, which violates the conditions of the questions

user254545
  • 83