Children And Toys Distribution Problem?

So you want ALL of the 5 toys to be distributed?
There are two different ways this could happen:
- one child gets 3 toys, and the others get one toy each
- two children get 2 toys each, and the remaining child gets one
Consider the first case. Let’s assume the first child is the one who gets the 3 toys. Then the child could have any of the first 5, then any of the reamaining 4, then any of the remaining 3 for a total of 5*4*3 / 3! = 10 different combinations. (We have to divide by 3! to take care of the fact that having toys “A, B and C” is the same as having toys “B, A, and C”, etc.). The remaining two toys can only be distributed two different ways between the remaining two children. So in total that’s 20 different combinations if the first child gets 3 toys. For all 3 children, it would be 3*20 = 60.
Now consider the second case. The first child gets any one of 5 toys. The second child gets one of the remaining 4 and then one of the remaining 3, making 4*3 / 2! = 6 different combinations. The remaining 2 toys have to go to the third child by default. This makes for 5 * 6 = 30 combinations if the first child gets stuck with one toy. So for all three, that’s 3*30 = 90.
Adding up all scenarios, we get 150.
That’s one way of calculating this. We could have tried a different approach, but we have to be careful not to count duplicate permutations.

There are 2 ways to distribute the toys , the toys that each child gets are : 3,1,1 or 2,2,1
So :
10*3*2 + 10*3*3*2 = 60+180 = 240