At the Party:

- There were 9 men and children.
- There were 2 more women than children
- The number of different man-woman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 man-woman couples possible.

Also, of the three groups - men, women and children - at the party:

- There were 4 of one group.
- There were 6 of one group.
- There were 8 of one group.

Exactly one of the above 6 statements is false.

Which one of the above statements is false? Also, how many men, women and children are there at the party?

Assume that one of the first three statements is false. Then, the three groups must be 4, 6, and 8, so the 1st statement is false. Therefore, the 2nd and 3rd are true. The number of couples is 24, so the men and woman must be 4 and 6 (in some order), making the children 8, making statement 2 false. Our initial assumption is wrong, and all of the 1st three statements are true.

Using the 1st two statements, the possibilities are only 10:

M C W Couples

9 0 2
**8 ** 1 3 24
7 2 4
6 3 5
5 4 6
4 5 7
3 **6 ** **8** 24 <== Solution
2 7 9
1 8 10
0 9 11

Only 2 of those possibilities are consistent with 24 possible couples, and only one of those use two of 4, 6, and 8.

So, the 4th statement is false and the solution is unique.

*Edited on ***December 25, 2012, 11:51 am**