Answer to Question #142113 in Discrete Mathematics for Promise Omiponle

Question

How many solutions are there to the equation x1+x2+x3+x4+x5+x6= 25 where the xi(for i= 1,2,3,4,5,6) are non-negative integers?

Answer 1

A k-combination with repetitions, or multisubset of size $k$  from a set $S$  is given by a sequence of $k$  not necessarily distinct elements of $S$, where order is not taken into account: two sequences define the same multiset if one can be obtained from the other by permuting the terms. Associate an index to each element of $S$  and think of the elements of $S$  as types of objects, then we can let ${\displaystyle x_{i}}$ denote the number of elements of type $i$ in a multisubset. The number of multisubsets of size $k$  is then the number of nonnegative integer solutions of the Diophantine equation:

$x_1+x_2+...+x_n=k$

If $S$  has $n$  elements, the number of such $k$-multisubsets is denoted by $({n \choose k})$. This expression, $n$  multichoose $k$, can also be given in terms of binomial coefficients:

$({n \choose k})={n+k-1\choose k}$.

In our case, the number of nonnegative solutions of the equation

$x_1+x_2+x_3+x_4+x_5+x_6= 25$

is equal to

$({6 \choose 25})={30\choose 25}=\frac{30!}{25!\cdot5!}=\frac{30\cdot 29\cdot 28\cdot 27\cdot 26}{5\cdot 4\cdot 3\cdot 2}=142,506$