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Official Solution:
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
Approach 1:
Total number of handshakes possible between \(10+7=17\) people (with no restrictions) is the number of different groups of two we can pick from these \(10+7=17\) people (one handshake per pair), so \(C^2_{17}\). The same way: # of handshakes between chairmen \(C^2_{7}\) (restriction).
\(\text{Desired}=\text{Total}-\text{Restriction}=C^2_{17}-C^2_{7}=115\).
Approach 2:
Direct way: number of handshakes between executives \(C^2_{10}\) plus \(10*7\) (as each executive shakes the hand of each 7 chairmen): \(C^2_{10}+10*7=115\).
Answer: C
Number of handshakes for first business executive = 16 ( Business executive one shakes hand with 9 other business executives and 7 chairmen)
Number of handshakes for second business executive =15 ( the handshake with the first executive is excluded here since it has been already counted for first business executive)
Similarly, Number of handshakes for tenth business executive= 7 ( the handshakes with other 9 executives are excluded here since those have already been counted)
Total number of handshakes = 10/2 *(16+7) = 5*23 = 115 (Sum of AP= total number of terms/2 * (first term+last term) )
Answer C
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