Question

a man has 8 children to take them zoo. he takes three of them at a time to the zoo as often as he can without taking the same 3 children together more than once. how many times will he have to go to the zoo? how many times a particular child will go to the zoo

Answer 1

If I understand correctly, then this is a combinations problem. Let's pretend he only has 4 children, A, B, C, and D: Then he might do this:

ABC
ABD
ACD
BCD

Now if he takes three kids, they'll be the same as one of those above, maybe in a different order.

Let's try 5:
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE.

Okay, you get the idea. In any case, you should recognize that what we're really doing is choosing how many ways you can divide the 8 children into groups of 3. You should recognize those words as meaning

8 choose 3 = 8!/3!5! = 56.

But that's not what the question asks. (56 is the answer to "How many times does the father go?")

First, every child goes the same number of times. That should be clear. So it really suffices to ask how many times does A go?
Well, notice how I made the charts above. I wrote all the ones with A first. How many of them did I write? Well, there were 3 people in each group, so I had to choose 3-1=2 more, and I couldn't choose A, so I chose 4-1=3 or 5-1=4.

So if there are 8 children, A would be in

(8-1) choose (3-1) = 7 choose 2 = 7!/5!2! = 21 groups,

so he would go 21 times.

answer is 56 and 21