At a party each man dances with 4 women and each woman dances with 3 men. If 9 men attended the party, how many women attended the party?

At a party each man dances with 4 women and each woman dances with 3 men. If 9 men attended the party, how many women attended the party?

I have no idea how to approach this problem.

• How many ways can a women choose 3 men in a set of 9 ? that would help. – user451844 Jul 13 '17 at 13:22

If you assume that a man dances with a woman, how many dances will be performed?

9 men -> dance 9 * 4 times with a woman, so 36 dances

Each woman performs 3 dances, so 12 women.

Count dance pairs . . .

Since there are $9$ men and each man dances with $4$ women (presumably this means each man dances exactly $4$ times), there are exactly $36$ dance pairs.

Let $w$ be the number of women.

Since each woman dances with $3$ men (presumably this means each woman dances exactly $3$ times), there are exactly $3w$ dance pairs.

Thus, $3w=36$, so $w=12$.

• Obviously, we agree, but I don't think once each really matters. 9 women can dance 3 times with the same men, and the men then perform 12 dances with the 3 leftovers. (But that would be a strange scheme) – Pieter21 Jul 13 '17 at 13:30
• The issue is if each man dances with 4 different women, but some men dance more than once with some of them. In that case, the number of dance pairs is not determined. – quasi Jul 13 '17 at 13:41
• As long as the condition "each man dances with 4 women" means each man dances exactly $4$ times. and similarly, if the condition "each woman dances with $3$ men" means each woman dances exactly $3$ times, then the choice of partnerships (as long as it's always (man, woman)) doesn't matter, and repeats are OK. I edited in this clarification. – quasi Jul 13 '17 at 13:54

Divide the $9$ men into $3$ groups of $3$. Assign $4$ women to each group. That'll satisfy the conditions of the problem, with a total of $12$ women.

First naive thought: each of the $9$ men dances with $4$ women, so we need $36$ women for that.

No, wait... Every woman danced with $3$ men, so stands in for $3$ of those $36$ women.

Then exactly $12$ women are needed.

I have a different view. It is given that one man danced with 4 women. W¹,W²,W³,W⁴-M¹ W²,W³,W⁴, W5-M² ' ' ' W1,W²,W5,W6-M9 This gives us a total of 6 women. This also satisfies that each woman danced with 3 men. How is it wrong?

• How does this show us each of the women dance with $3$ men? – Daniel Buck Aug 11 '18 at 17:30