The Smiths, the Andrings and the Cliffords all hold a big party. Everyone shakes hands with every member of the other two families (no one shakes hands with members of their own family), 142 handshakes in all.

Assuming that there at least as many Andrings as Smiths, and at least as many Cliffords as Andrings, how many of each family are present?

(In reply to Solution by e.g.)

Under the assumption that 1 person can make up a family, one possible solution is:
1 Smiths, 10 Andrings, and 12 Cliffords.

Two additional possibilities exist if there were actually 0 Smiths present at the Smiths-Andrings-Cliffords gala:
0 Smiths, 1 Andrings, and 142 Cliffords; and
0 Smiths, 2 Andrings, and  71 Cliffords,  

Posted by Dej Mar on 2008-11-03 21:58:15