Question

A father with 8 children takes them 3 at a time to the zoological garden, as often as he can without taking the same three children together more than once. Then,
(A) Number of times he will go is 56
(B) Number of times each child will go is 21
(C) Number of times each child will go is 35
(D) Number of times a child will not go is 35.

Answer 1

Hint: We start solving this question by first finding the number of times the father will go to the garden using the formula for selection of r objects from n, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$. Then we find the number of times a child can go to the garden using the same formula. Then we subtract the number of times a child can go from the total number of times father goes to find the number of times a child does not go.

Complete step-by-step answer:
We are given that the number of children is 8 and the father takes them to a zoological garden as often as he can without taking the same 3 children together more than once.
The number of times he will go to the garden is equal to the number of ways of selecting three children from a total of the 8 children.
Now let us use the formula for number of ways of selecting r object from a set of n objects,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$
So, applying the above formula we get that the number of ways of selecting 3 children out of 8 children is
${}^{8}{{C}_{3}}=\dfrac{8!}{3!\times 5!}=\dfrac{6\times 7\times 8}{6}=56$
So, the number of times he can go to the garden is 56.
Now, let us find the number of times a child can go to the garden.
A child can go to the garden until he cannot go with the same two other children.
So, number of times a child goes is equal to number of ways of selecting 2 children out of 7, that is equal to
${}^{7}{{C}_{2}}=\dfrac{7!}{2!\times 5!}=\dfrac{6\times 7}{2}=21$
So, each child will go to the garden 21 times.
The total number of times father goes to the garden is 56 and each child goes is 21. Number of times a child does not go to the garden is $56-21=35$ times.
So, finally we get that,
Number of times father will go to the garden is 56.
Number of times each child goes is 21.
Number of times a child will not go is 35.

So, the correct answers are “Option A,B and D”.

Note: The common mistake that one does while solving this type of problem is one might take the formula for number of ways of selecting r object from a set of n objects wrong as, ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. One might also stop finding the values of the number of times a child goes and does not go after finding the number of times the father goes and seeing it as option A. But we need to find the values as the question might have multiple answers.