Number of handshakes - exclusion apporach

Question

5 indian and 5 american couples meet at a party and shake hands. If no wife shakes hands with her husband and no indian wife shakes hands with a male, then the number of hand shakes that take place at the party is?

I know this question has been asked before here and I understand the solution but I can't figure out what am I missing in my approach.

Assumption - Everyone shakes hands with everyone they can according to these restrictions, and that no one shakes hands with the same person twice.

Here's what I did -

1. Total number of handshakes by Indian females: $ {10 \choose 2}$ since they can only shake hands with the american wives.
2. The total number of handshakes by the other fifteen people will be: $ {15 \choose 2}$
3. Subtract from that number of handshakes by american wives with their own husband: 5

So the answer should be $ {10 \choose 2}$ + $ {15 \choose 2}$ - 5 = 145

But the answer is 135. Can somebody point out what I am I missing in this approach?

Answer 1

When you calculate the number of handshakes done by the Indian wives, $\binom{10}2$ also picks up the handshakes done between all the American wives. There are ten of those.