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A father with 10 children takes those 3 at a time to the zoo, as often as he can without taking the same 3 children together more than once. Then

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The number of times each child will go is 36

Number of times a particular child will not go is 84

The number of times he will go is 120.

At the above

answer is A, B, C, D.

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Detailed Solution

No. of ways selecting 3 form 10= ${}^{10}C_{3} = \frac{10.9.8}{1.2.3} = 120$ , $^{9} C_{2} = 36, 120 - 36 = 84$

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