P&C

[Captaingmat]

2 men, 3 women and 5 chairs. How many ways if no two men sit together?

I dont know the answer. I got 8 ways.

[ganeshrkamath]

2 men, 3 women and 5 chairs. How many ways if no two men sit together?

I dont know the answer. I got 8 ways.

Men: A,B
Women: P,Q,R

Number of ways in which A and B sit together:
AB _ _ _
_ AB _ _
_ _ AB _
_ _ _ AB
Total = 4*2*3! = 48

Total number of possible arrangements = 5! = 120

So the number of ways in which the two men do not sit together = 120 - 48
= 72

Cheers

[GMATGuruNY]

2 men, 3 women and 5 chairs. How many ways if no two men sit together?

I dont know the answer. I got 8 ways.

Let A and B = the 2 men and C, D, and E = the 3 women.

Good arrangements = total arrangements - bad arrangements.

Total arrangements:
Number of ways to arrange the 5 elements A, B, C, D and E = 5! = 120.

Bad arrangements:
In a BAD arrangement, A and B are in adjacent positions.
Put A and B together in a BLOCK, so that AB serves as SINGLE ELEMENT in the arrangement.
Number of ways to arrange the 4 elements AB, C, D, and E = 4! = 24.
in each of these arrangements, AB can be reversed to BA.
Thus, we multiply by 2:
2*24 = 48.

Good arrangements:
Total-bad = 120-48 = 72.