TITLE: Determine number of ways to distribute sixteen identical pieces of candy to five non-identical kids...
QUESTION [5 upvotes]: Provided that the youngest kid must receive no more pieces than any of the others. All kids are of different ages.
Just want to make sure my logic is correct...
So using the pigeonhole principle, one kid will have at least 4 pieces, because 16/5 = 3 remainder 1
With that in mind, that means that the youngest kid must be getting 3 pieces at most.
I take it that it makes sense to first assign 0, 1, 2, or 3 to each then go about assigning the rest. That said, there will be 4 cases:
1) Give each 0, assign 16 to other 4:
C(20, 16)
2) Give each 1, assign 11 to other 4:
C(15,11)
3) Give each 2, assign 6 to other 4:
C(10,6)
4) Give each 3, assign 1 to other 4:
C(5,1)
Then we would add them all up...
Will this lead to the correct answer?

REPLY [3 votes]: Let us give the youngest child $0,1,2,3$ candies and the same number to the other four (because they can't get less)
So the problem reduces to distributing $16, 11, 6, 1$ candies any which way to the other $4$ children
Let us consider, e.g. $6$ candies $\;\bullet\bullet\bullet\bullet\bullet\bullet\;$ to be distributed.
To divide between  $4$ children, you need to draw three dividers.
A few illustrative examples are:
$\;\bullet\bullet|\bullet|\bullet\bullet|\bullet\;\;$ representing a distribution of $\;2-1-2-1$
$\;\bullet|\bullet\bullet\bullet\bullet|\bullet|\;\;$ representing a distribution of $\;1-4-1-0$
$\;|\bullet\bullet|\bullet\bullet\bullet|\bullet\;\;$ representing a distribution of $\;0-2-3-1$ and so on.
In other words, we have to place the $3$ dividers in the lot of $9$ symbols in $\binom93$ ways
Similarly for the other $3$ cases, so ans = $\binom{19}{3} + \binom{14}3 + \binom93 +\binom43 = 1421\;$ ways.
The technique is known as "stars and bars", and you can get a more formal explanation of it here