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 $3$ children together more than once. Then the number of times a particular child will not go to the zoological garden is

Answer 1

Given total number of children's$=8$ and father of them take $3$ children's at a time to the zoological garden, as often as he can without taking the same $3$ children.

So, the number of times that the father will go to the garden is equal to the number of ways of selecting three children from a total of the $8$ children.

This can be done by ${}^8{C}_3=\frac{8!}{5!\times 3!}=\frac{8\times 7\times 6\times 5!}{5!\times 3\times 2\times 1}=56$ ways.

Therefore, the number of times he can go to the garden is $56$ times.

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 ${}^7{C}_2=\frac{7!}{2!\times 5!}=\frac{7\times 6\times 5!}{1\times 2\times 5!}=21$

Therefore, number of times each child will go to the garden is $21$ times.

Thus, number of times a child does not go to the garden = total number of times father goes to the garden$-$ number of each child will go to the garden.

$=56-21$

$=35$

Hence, the number of times a child will not go is $35$.