A father with 6 children takes 3 of them  at a time to the zoo, as often as he can without taking the same three children together more than once. How often will he go? How often will each child go ?

The problem can be solved using the concept of combinations in mathematics.

A combination is a selection of items without considering the order. In this case, the father has 6 children and he takes 3 at a time to the zoo. We want to find out how many different groups of 3 can be selected from 6.

The formula for combinations is:

C(n, r) = n! / [(n-r)! * r!]

where:

• n is the total number of items,
• r is the number of items to choose,
• "!" denotes factorial, which is the product of all positive integers up to that number.

Substituting the given values into the formula:

C(6, 3) = 6! / [(6-3)! * 3!] = (654321) / [(321) * (32*1)] = 720 / 36 = 20

So, the father will go to the zoo 20 times.

Now, to find out how often each child will go, we consider that each group of 3 children includes a particular child. So we want to find out how many groups of 2 can be selected from the remaining 5 children (since one child is already considered).

Using the combinations formula again:

C(5, 2) = 5! / [(5-2)! * 2!] = (54321) / [(321) * (2*1)] = 120 / 12 = 10

So, each child will go to the zoo 10 times.