How many ways are there to distribute $3$ different teddy bears and $9$ identical lollipops to four children:

(a) without restriction?

(b) with no child getting greater than or equal to $2$ teddy bears?

(c) with each child getting only $3$ goodies?


My work/attempts:

I'm pretty confident with my first two responses...

(a) Since the teddy bears are distinct there are $4^3$ ways to distribute them to four children. Meanwhile, since the lollipops are identical, we can consider them as dots and the $4$ children having $3$ slashes to separate them. We then can use the choose function to determine the number of ways to distribute them: ${9+3\choose 3}$ = ${12\choose 3}$ ways. Therefore there are $4^3{12\choose 3}$ outcomes.

(b) Since the children can not get greater than or equal to $2$ bears we will assume there is no repetition of the bears. Thus rather than having $4^3$ ways to distribute the distinct bears to the 4 children, there will be $4$ x $3$ x $2$ ways to distribute the bears (or 4!). Again the lollipops are identical so we still have ${12\choose 3}$ to distribute these to the 4 children. Thus our number of outcomes for this case is 4!${12\choose 3}$.

(c) I really am not sure how to go about this problem because I'm not sure how to work with mixing the types of goodies since one has the bears distinct and the other type has the lollipops identical.

  • $\begingroup$ If you find answer helpful then accept the answer $\checkmark$ :) $\endgroup$ – Key Flex Nov 4 '18 at 19:58

The key for (c) is that once the teddy bears are distributed, since the lollipops are identical and the size of each allocation is fixed there is only one way to distribute the lollipops. Thus, each teddy bear may be sent to one of the four children independently. (one child can receive all the teddy bears, leaving all the lollipops to others). The answer is $4^3=64$.


$1)$ There are $4^3$ ways to distribute the teddy bears and $\dbinom{9+4-1}{9}$ ways of distributing the lollipops.

In total there are $4^3\times\dbinom{9+4-1}{9}$ ways without restriction.

$2)$ Since each can have at most one teddy bear, so one child must go without. The teddy bears can be distributed to $3$ children in $3!$ ways. We can distribute the lollipops in $\dbinom{9+4-1}{9}$ ways. There are $4$ ways to choose a child who will not receive a teddy bear. In total there are $4\times3!\times\dbinom{9+4-1}{9}$ ways.

$3)$ We can distribute $3$ teddy bears to $4$ children in $4^3$ ways.

We can ensure that each child gets $3$ goodies by filling out each child with lollipops.

So, there are a total of $4^3=64$ ways


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