Here's the question which I'm stuck with -

There are $20$ married couples at a party. Every man shakes hands with everyone except himself and his spouse. Half of the women refuse to shake hands with any other women. The other $10$ women all shake hands with each other (but not with themselves). How many handshakes are there at the party?

My Solution:

  1. Handshakes done by men: There are $20$ ways to pick a man and $38$ ways to pick the other person, totaling to $20 \cdot 38 = 760$. But since every handshake is counted twice, the answer is $\frac{760}{2} = 380$
  2. Handshakes done by the women who refuse to shake hand with any other women: these are already counted in $380$ handshakes (Because the women can only shake hands with men. This was taken care of above).
  3. Handshakes done by the other $10$ women and men: These are counted in $380$ handshakes.
  4. Handshakes done by women and women in the group of $10$: are $\binom{10}{2} = 45$.

Hence the total handshakes are $380 + 45 = 425$.

However, the total handshakes are $615$ according to my textbook. Can anyone please help me to find out the mistake?

  • 2
    $\begingroup$ After you compute the number of handshakes done by each man, you divide by 2 because "every handshake is counted twice". But, you also ignore handshakes between men and women because those are "already counted in [the] 380 handshakes". This means you're miscounting something here -- you're missing a whole bunch of handshakes between men and women. I recommend waiting until the very end to divide anything by 2. $\endgroup$ – nukeguy Feb 21 '16 at 7:10
  • $\begingroup$ There is a non-maths problem in here: this question is unanswerable unless you assume that each married couple consists of one man and one woman, which is no longer a valid assumption these days and is not stated in the question. $\endgroup$ – hvd Feb 21 '16 at 14:05
  • $\begingroup$ Good problem. While reading it, I did the exact same reason you did for some reason. Lol ! $\endgroup$ – user230452 Feb 21 '16 at 14:21

Your mistake can be seen in your first line: you should not divide by $2$ as you did not count the handshakes between men and women twice.

Instead, the ways to pick a man is $20$. The number of men that shake hands with him is $19$. Since very handshake is counted twice, the men shake hands $190$ times.

The number of handshakes men had with women is simply $20 \times 19$, thus $380$.

Combining this with your results gives the answer in the textbook, or $570+45=615$.


There are the following types of handshakes

  1. Woman vs woman: $\dbinom{10}{2}=45$. You were correct on that.
  2. Man vs man: $\dbinom{20}{2}=190$.
  3. Man vs woman: each man performs $19$ handshakes with women. Since there are $20$ men this gives $$19\times20=380$$ such handshakes.

Summing up $$45+190+380=615$$ Your mistake was in steps $2$ and $3$ because you treated them as $1$ step. While handshakes between man and man are counted indeed twice, handshakes between man and woman are counted only once, so dividing all with $2$ is wrong. You have to distinguish between these two categories.


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