5 Indian and 5 American couples meet at a party and shake hands. If no wife shakes hands with her own husband and no Indian wife shakes hands with a male, then find the number of hand shakes that took place in the party.

Since there are 5 Indian couples and 5 American couples, so there are a total of 20 persons in the party. So, the total number of handshakes possible = ²⁰C₂ It comes out to be 190 (20!/(2!*18!). Now if no wife shakes hand with her own husband, then = 190-10=180 It is also given that no Indian wife shakes hand with any male and there are 5 Indian wives.  Any of them cannot shake hand with 10 males in the party, but we have already included the case in which they do not shake hands with their own husband. So, each of them cannot shake hands with 9 males in the party So, the number of rejected handshakes = 5*9 = 45. Total number of handshakes = 180-45 = 135 (Correct Answer)