Determine number of ways to distribute sixteen identical pieces of candy to five non-identical kids... 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?
 A: 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
A: The youngest kid is fixed. Hence as you say he cannot receive more than three pieces. We break it into cases as you said, but I want to point out a small flaw in your argument.
Case 1: The youngest kid receives  $3$ pieces. So there are $13$ pieces left for the other four kids, but note that each kid must receive at least $3$ pieces. Hence $13-12=1$ piece is left over after you give three to each of the other four kids, but this one piece can be given to any one of them, and that can be done in $4^C_1=4$ ways.
Case 2: The youngest kid receives  $2$ pieces. So there are $14$ pieces left for the other four kids, but note that each kid must receive at least $2$ pieces. Hence $14-8=6$ pieces are left over after you give two to each of the other four kids, and now you have to find the number of ways you can distribute $6$ candies to $4$ kids. For this, draw six dots like this:
$$
......
$$
Note that by placing 3 bars in between the dots, we separate them into different partitions of dots:
$$
..|..|.|.
$$
The above would correspond to the first two children getting two more candies and the next two getting one more each. Similarly,
$$
|.....||.
$$
would correspond to the first and third kids getting no more candies, the second child getting $5$ more candies and the fourth one getting $1$ more candy.
Thus, the number of ways of distributing candies is the number of ways of putting $3$ bars in $9$ slots,because the total number of bars and dots is $9$ and you have to decide which are the three bars and the six dots, the answer to which is $^9C_3 = 84$.
Case 3: The youngest kid receives  $1$ piece. So there are $15$ pieces left for the other four kids, but note that each kid must receive at least $1$ pieces. Hence $15-4=11$ pieces are left over after you give one to each of the other four kids, and using a similar logic to above, the remaining  11 chocolates can be given to the $4$ children in $^{14}C_3=364$ ways.
Case 4: The youngest kid receives no pieces. So there are $16$ pieces left for the other four kids, and using a similar logic to above, the $16$ chocolates can be given to the $4$ children in $^{19}C_3=969$ ways.
The total then comes out to be $1421$. You should think about what happens if you fix two children,and if two children are,well, identical twins.
