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08 Dec 2021, 06:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Chris and his wife invited a total of ten families on their marriage anniversary. While the host family had just the two members, each family invited consisted of four members. If every person in the party shook hands exactly once with every other person belonging to a different family, then find the total number of handshakes that took place in the party.
(A) 729
(B) 800
(C) 801
(D) 850
(E) 851
Are You Up For the Challenge: 700 Level Questions
(A) 729
(B) 800
(C) 801
(D) 850
(E) 851
Are You Up For the Challenge: 700 Level Questions
08 Dec 2021, 07:10
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We have ten families of 4, and one family of 2, so there are 42 people in total. If everyone shook hands with everyone else, then every pair of people would produce a handshake, and we have 42C2 = (42)(41)/2! = 861 pairs of people. So if there were no restrictions, the answer would be 861. But we know that within each family, no one shakes hands. In the two-person family, that means 1 handshake doesn't take place, and within each of the ten four-person families, 4C2 = 6 handshakes don't take place. So there are 1 + 60 = 61 handshakes we shouldn't count, and the answer is 861 - 61 = 800.
Or you could do the problem directly. First, just focusing on the families of four (ignoring the family of two completely for now), we have 40 people in total. Each of these 40 people will shake 36 hands, among only the families of four, so within the families of four, we'll have 40*36/2 handshakes (dividing by 2 because we don't want to count each handshake twice, once for each person involved). So among the families of four, we have 720 handshakes in total. But the two hosts shake everyone else's hand, so they're each involved in an additional 40 handshakes, and the answer is 720 + 40 + 40 = 800.
Or you could do the problem directly. First, just focusing on the families of four (ignoring the family of two completely for now), we have 40 people in total. Each of these 40 people will shake 36 hands, among only the families of four, so within the families of four, we'll have 40*36/2 handshakes (dividing by 2 because we don't want to count each handshake twice, once for each person involved). So among the families of four, we have 720 handshakes in total. But the two hosts shake everyone else's hand, so they're each involved in an additional 40 handshakes, and the answer is 720 + 40 + 40 = 800.
08 Dec 2021, 09:28
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Bunuel
Ten families of 4 members each + Host family of 2 members = 42 Total members
Now if everyone handshakes everyone we get = 42C2 = 861 handshakes
We have to remove the handshakes that each family did amongst themselves.
10C1 x 4C2 (handshakes for the 10 families) + 1 (host family handshake) = 61
Residual handshakes = 861 - 61 = 800
Option B
