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A father takes 8 children, 3 at a time to the zero as often as he can take without the same three together more than once. If the number of times each children can go is x and the number of times he can go is y, then (x, y)=

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(21, 53)

(20, 56)

(20, 53)

(21, 56)

answer is D.

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Detailed Solution

x = The number of times each children can go $= 7_{C_{2}} = 21$

y = The number of times the father can go $= 8_{C_{3}} = 56$

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