Rolling 10 dice.

We want the coefficient of $x^{15}$ in $\big(x + x^2 + x^3 + \cdots + x^{20}\big)^{10} \; = \; x^{10}\big(1 + x + x^2 + \cdots + x^{19}\big)^{10} \; = \; x^{10}\left(\frac{1 - x^{20}}{1-x}\right)^{10}$ which is therefore the coefficient of $x^5$ in $(1-x)^{-10}$ , which is $\binom{14}{9} = \boxed{2002}$ .

Consider all arrangements of $15-10=5$ $X$ s and $10-1=9$ $I$ s, such as $IIXIIIIXIIXXIX.$ We can bring these into a $1$ to $1$ correspondence to the possible ways of throwing our $10$ dice. Each $I$ marks the beginning of the next dice and we count the amount of $X$ s in between the $I$ s and add $1$ for the amount rolled. The example is thus put into a correspondence with $(1,\: 1,\: 2,\: 1,\: 1,\: 1,\: 2,\: 1,\: 3,\: 2)$ . The amount of ways to put $9$ $I$ s in between (or around) the $5$ $X$ s is ${{5+9} \choose {9}} = \boxed{2002}$ .

Therefore, in general, for $n$ dice with each $p$ faces and a total sum of $k \leq p$ , we have a total amount of $k-1 \choose n-1$ possible combinations.

$15 = 9(1) + 1(6) \\ \;\;\;\; = 8(1) + 1(2) + 1(5) \\ \;\;\;\; = 8(1) + 1(3) + 1(4) \\ \;\;\;\; = 7(1) + 2(2) + 1(4) \\ \;\;\;\; = 7(1) + 1(2) + 2(3) \\ \;\;\;\; = 6(1) + 3(2) + 1(3) \\ \;\;\;\; = 5(1) + 5(2)$

The number of ways to roll a total of $15$ is: $10!\left( \dfrac{1}{9!} + \dfrac{1}{8!} + \dfrac{1}{8!} + \dfrac{1}{7!2!} + \dfrac{1}{7!2!} + \dfrac{1}{6!3!} + \dfrac{1}{5!5!} \right) = 2002$

I'm not proud of the code I wrote for this but here it is
Only thing I can say is with rolling 10 dice, minimum of 1 per dice, that means youre never going to want any number over 6, so we can save ourselves some time on loops by not counting to 20 and by using the 'continue' keyword any time our sum exceeds 15.

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<?php
$s = 0;
for($a=1;$a<=6;$a++){
    for($b=1;$b<=6;$b++){
        for($c=1;$c<=6;$c++){
            for($d=1;$d<=6;$d++){
                for($e=1;$e<=6;$e++){
                    for($f=1;$f<=6;$f++){
                        for($g=1;$g<=6;$g++){
                            for($h=1;$h<=6;$h++){
                                for($i=1;$i<=6;$i++){
                                    for($j=1;$j<=6;$j++){
                                        $sum = array_sum(array($a,$b,$c,$d,$e,$f,$g,$h,$i,$j));
                                        if($sum==15){
                                            $s++;
                                        } elseif($sum>15){
                                            continue;
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}
echo $s;
?>

There is a formula to find how many ways are there to get sum ${p}$ when rolling $n$ dice each having $s$ faces:

$\sum^{k_{max}}_{k=0} (-1)^{k} {n \choose k} {p - s \times k - 1 \choose p - s \times k - n}$

where $k_{max} =$ floor of $\frac{p - n}{s}$

In this case its ${10 \choose 0} {15 - 1 \choose 15 - 10} = {14 \choose 5} = 2002.$

$a_{1} +a_{2} +a_{3} + ... + a_{10} = 15$ ( $1 ≤ a_{i}$ )

Let $a_{i} -1 = b_{i}$ and then,

$b_{1} +b_{2} +b_{3} + ... + b_{10} = 5$ with ( $0 ≤ b_{i}$ )

So, we get 10H5 = 14C5 = 2002