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

OA: C

Chairmen shake hands 10*7=70 with business executives times. Business executives shake hands with each other 10 C 2 times or 45 times. The total is 115 .

**Why 10C2?**10 business executives shakes hands with other 9 business executives in 10c2 ways = 45 ways

10c2 = 9+8+7+6+5+4+3+2+1 = 45

First executive shakes hands with remaining 9 executives

Second executive shakes hands with remaining 8 executives

Third executive shakes hands with remaining 7 executives

Fourth executive shakes hands with remaining 6 executives

Fifth executive shakes hands with remaining 5 executives

Sixth executive shakes hands with remaining 4 executives

Seventh executive shakes hands with remaining 3 executives

Eighth executive shakes hands with remaining 2 executives

Nineth executive shakes hands with remaining 1 executives

Tenth executive already shakes hands with all 9 executives.

## 2 comments:

1. There are 10 BE and 7 CM.

2. Now if everyone shakes hands with everyone then there will be

17C2 handshakes.

3. If we only consider the CMs shaking hands with each other then there would have been 7C2 handshakes.

4. In the total handshakes we calculated above the handshakes between CMs are also included, so we'll subtract that number.

= (17C2 - 7C2)

= 115 handshakes

PS: (-:-D

:-( again, didn't want another method to solve the question. My doubt remains...why 10C2?

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