NandishSS wrote:10 business executives and 7 chairmen meet at a conference. If each business executive shakes the hand of every other business executive and every chairman once, and each chairman shakes the hand of each of the business executives but not the other chairmen, how many handshakes would take place?
A. 144
B. 131
C. 115
D. 90
E. 45
Recall that since it takes 2 people to shake hands with each other, the number of handshakes that can be made in a room of n people is nC2.
Now if we assume that all 17 people can shake hands with each other, then the number of handshakes that can be made is 17C2 = 17!/[2!(17-2)!] = (17 x 16)/2 = 17 x 8 = 136.
The number 136 includes the handshakes among the chairmen. However, Since the chairmen can't shake hands with each other, we need to subtract the number of handshakes among the chairmen from 136.
The number of handshakes among chairmen is 7C2 = 7!/[2!(7-2)] = (7 x 6)/2 = 7 x 3 = 21.
Thus the total number of handshakes, if no chairmen shake hands with each other, is:
136 - 21 = 115
Alternate Solution:
Let's calculate the number of handshakes each of the business executives and chairmans make.
Since business executives shake the hands of every person besides himself/herself, each business executive will shake hands with 10 - 1 + 7 = 16 people. Since there are 10 business executives, the total will be 16 x 10 = 160.
Since a chairman only shakes hand with business executives, each chairman will shake hands with 10 people and since there are 7 chairman, the total will be 7 x 10 = 70.
Thus, the grand total is 160 + 70 = 230. However, each handshake is counted twice in this sum; therefore the number of handshakes that took place will be 230 / 2 = 115.
Answer:
C
