In how many ways can I distribute $6$ identical cookies and $6$ identical candies to $4$ children, if each child must receive at least $1$ of each type of item?

I know how to distribute the things if they are a specific amount, but I'm struggling with the "at least" part.

• I removed the probability-distributions tag; this has nothing to do with probabilities. Also, give your children more healthy food :-) – joriki Jun 23 '18 at 3:33
• Hint: start by giving them each 1 cookie and 1 candy, since that is required. Now the problem is simpler. – Théophile Jun 23 '18 at 3:41

Lets say $A,B,C,D$ are $4$ children and you need to distribute $6$ identical candies and $6$ identical cookies.

Now first each $A,B,C,D$ children will get $1$ candy and $1$ cookie.

So, now we have to distribute the remaining $2$ candies and $2$ cookies among $4$ children.

To distribute $2$ candies use the formula $$^{n+r-1}C_{n-1}$$ where $n=4,r=2$ $$^{4+2-1}C_{4-1}=\ ^ 5C_3=\frac{5!}{2!\cdot3!}=10\ ways$$

Again to distribute $2$ cookies use the formula $$^{n+r-1}C_{n-1}$$ where $n=4,r=2$ $$^{4+2-1}C_{4-1}=\ ^ 5C_3=\frac{5!}{2!\cdot3!}=10\ ways$$

So, total number of ways $=10\times 10=100$ ways.

One candy and one cookie can be given to one of the 4 children in 6*6 ways.i-e 36 ways.

Likewise second candy and second cookie can be given in 25 ways.

Similarly third and fourth candy and cookie can be given in 16 and 9 ways respectively.

Now 2 cookies and 2 candies remains. They can be distributed to any one of 4 children in 4 ways each i-e $4^4 ways$.

Hence the total number of ways of distributing 6 cookies and 6 candies to 4 children so that each one of them get at least one cookie and one candy are 36*25*16*9*$4^4$ ways.