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: 
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.
> 
> 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 $$\binom{15+5-1}{5-1}$$