# 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?

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

• Are they sitting in a row or in a circle? And if in a circle, are arrangements equivalent if they differ only by a rotation? – joriki Aug 15 '15 at 12:13
• How do you come to $5$ places for the $3$ women in the second approach ? – Peter Aug 15 '15 at 12:13
• @joriki Edited the question – Jack Aug 15 '15 at 12:17
• @Jalaj Chawla you need to state how many seats are there at this table – Wojciech Karwacki Aug 15 '15 at 12:21
• @WojciechKarwacki 7 people 7 seats.And i found my mistake. – Jack Aug 15 '15 at 12:24

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