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16 Sep 2014, 00:35
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Difficulty:
45%
(medium)
Question Stats:
68% (02:02) correct
32%
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based on 637
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At a conference, there are 10 business executives and 7 chairmen. If each business executive shakes hands with every other executive and each chairman exactly once, and each chairman shakes hands only with the business executives (but not with other chairmen), how many handshakes take place?
A. 144
B. 131
C. 115
D. 90
E. 45
A. 144
B. 131
C. 115
D. 90
E. 45
16 Sep 2014, 00:35
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Bookmarks
Official Solution:
At a conference, there are 10 business executives and 7 chairmen. If each business executive shakes hands with every other executive and each chairman exactly once, and each chairman shakes hands only with the business executives (but not with other chairmen), how many handshakes take place?
A. 144
B. 131
C. 115
D. 90
E. 45
Approach 1:
The total number of possible handshakes between 10 business executives and 7 chairmen (without any restrictions) is equivalent to the number of unique pairs we can form from these 17 individuals, which is \(C^2_{17}\). Similarly, the number of handshakes just between the chairmen is \(C^2_{7}\) (restricted handshakes).
Thus, the desired total is: \(\text{Total handshakes} - \text{Restricted handshakes} = C^2_{17} - C^2_{7} = 136 - 21 = 115\).
Approach 2:
Another method involves directly calculating the number of handshakes between the executives, which is \(C^2_{10}\), and then adding the handshakes between the executives and the chairmen, which is \(10 * 7\) (as each of the 10 executives shakes the hand of each of the 7 chairmen). So, the total is: \(C^2_{10} + 10 * 7 = 115\).
Answer: C
At a conference, there are 10 business executives and 7 chairmen. If each business executive shakes hands with every other executive and each chairman exactly once, and each chairman shakes hands only with the business executives (but not with other chairmen), how many handshakes take place?
A. 144
B. 131
C. 115
D. 90
E. 45
Approach 1:
The total number of possible handshakes between 10 business executives and 7 chairmen (without any restrictions) is equivalent to the number of unique pairs we can form from these 17 individuals, which is \(C^2_{17}\). Similarly, the number of handshakes just between the chairmen is \(C^2_{7}\) (restricted handshakes).
Thus, the desired total is: \(\text{Total handshakes} - \text{Restricted handshakes} = C^2_{17} - C^2_{7} = 136 - 21 = 115\).
Approach 2:
Another method involves directly calculating the number of handshakes between the executives, which is \(C^2_{10}\), and then adding the handshakes between the executives and the chairmen, which is \(10 * 7\) (as each of the 10 executives shakes the hand of each of the 7 chairmen). So, the total is: \(C^2_{10} + 10 * 7 = 115\).
Answer: C
25 Feb 2016, 13:49
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I understood it the way it was intended but I can see how some people could be confused.
The way I solved was by multiplying execs by chairmen = 70
Then I thought, the first Exec can shake 9 other executives. Then the next one can shake 8, 7,6,5,4,3,2,1 which adds to 45.
70+45 = 115
The way I solved was by multiplying execs by chairmen = 70
Then I thought, the first Exec can shake 9 other executives. Then the next one can shake 8, 7,6,5,4,3,2,1 which adds to 45.
70+45 = 115
