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Chris Peckover
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Oct 13 to Jan 7, 2027

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Logan Thompson
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Sep 6 to Dec 6, 2026

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Expert replies
Source: — Problem Solving |

by ganeshrkamath » Mon Sep 30, 2013 3:19 am
Captaingmat wrote: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

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by GMATGuruNY » Mon Sep 30, 2013 4:28 am
Captaingmat wrote: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.
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