Sunday, June 14, 2009

Problem 48 - GMAT Combinatorics

How many different handshakes are possible if six girls are standing on a circle and each girl shakes hands with every other girl except the two girls standing next to her?

(A) 12
(B) 11
(C) 10
(D) 9
(E) 8


  1. Use the formula to find the number of diagonals of a polygon for this problem.

  2. I would say each girl will shake hands with three girls (6 girls minus herself and the two girls next to her) so you got 6 times 3 equals but then you gotta discount the fact that Jane shaking Mary's hand is the same as Mary shaking Jane's so in the end you got 6(3)/2 = 9. So choice D?

  3. Both Eduardo and Anonymous are correct, although Eduardo's way is not so easy to see. What Eduardo is saying is find all of the possible combinations of two girls (all the handshakes) and from that subtract all the handshakes of each girl with the girl to the right of her. So
    6!/(2!)(4!)= 15, and 15 -6 = 9. However I see anonymous solution easier.

  4. Assume A B C D E F are seated on a circular table. A can shake hand with CDE( so per head 3 people and as per the restriction B and F not allowed) , so total 6 x 3 = 18 hand shakes.

    But try to put some couple of alternatives like
    E shakes hand with A B C
    B shakes hand with D E F

    You can clearly see we have counted it twice.....thus the total hand shake are 18 / 2 = 9