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:
- 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 $\frac{760}{2} = 380$
- 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).
- Handshakes done by the other $10$ women and men: These are counted in $380$ handshakes.
- 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?
Best Answer
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 $570+45=615$.