In how many ways can we give $20$ similar shirts to $5$ students?

Question

In how many ways can we give $20$ similar shirts to $5$ students so that the first gets at least $3$ shirts and the second gets at least $2$ shirts.

I thought the following:

$$\binom{20}{3}\cdot \binom{20-3}{2}\cdot (20-3-2)!$$ Is this correct?

Answer 1

Give 3 shirts to 1st student and 2 shirts to 2nd student & you are left with 15 shirts to be distributed between 5 students.

You can distribute it in $\binom {19}4$ ways.

Answer 2

The $2$ and $3$ are a distraction. One way to think about the problem is to assign those $5$ shirts at the beginning and distribute the remaining $15$ shirts. Then every student is treated the same.

As to the rest of the problem, assigning $15$ similar shirts to $5$ dissimilar students, this question is effectively asking how many shirts each student gets. One way to deal with this can be called stars and bars counting. We call the $15$ shirts stars and the $5$ students bars, and we arrange them in a line so that the right-most character is a bar, e.g. $$*****|***|**|**|***|$$ This groups the stars into $5$ groups representing how many shirts each student gets. The number of ways to distribute the shirts is then the same as the number of ways of creating such stars and bars diagrams. The right bar is fixed, and the remaining $19$ characters can move anywhere. There are only two types of characters, so an arrangement is obtained simply by choosing where the bars go. There are $4$ non-fixed bars, so your final answer is $$19\choose4$$

To answer your question completely, this is much smaller than the original number you arrived at.

Answer 3

In how many ways can we give $20$ similar shirts to $5$ students so that the first gets at least $3$ shirts and the second gets at least $2$ shirts.

Here is a very easy approach to think of...

We start off by giving the first person $3$, then $2$ shirts to the second one and we would have $20-(3+2) = 15$ shirts left to be distributed to the $5$ children under no restrictions.

Let xi be the number of shirts the ith child will be given for i belongs to {$1,2,3,4,5$}

$x1,x2,x3,x4,x5 = 20, xi>= 0$ ( Non-negative integers)

By using the stars and bars method, we have

$$\binom{k+n-1}{n-1}$$

Where k is the number of remaining shirts = $15$ and n is the number of people = $5$ so it would be $$\binom{15+5-1}{5-1}$$