Number of terms in the expansion

Question

The number of terms in the expansion of $$\left(x+y+z+w\right)^{10}$$

Answer 1

Each term is of the form $x^a y^b z^c w^d$ with $a+b+c+d=10$.

We have four elements, $x,y,z,w$, and we need to choose $10$ with repetition ignoring order.

This is called combination with repetition, or "multiset coefficient".

The required number is $\left(\dbinom{4}{10}\right) = \dbinom{4+10-1}{4-1} = \dbinom{13}3 = 286$.

Answer 2

We can start by considering smaller values for the exponent and seeing if we notice a pattern:

\begin{align}(x+y+z+w)^1&\text{ has } 4\text{ terms}\\ (x+y+z+w)^2&\text{ has } 10\text{ terms}\\ (x+y+z+w)^3&\text{ has } 20\text{ terms}\\ (x+y+z+w)^4&\text{ has } 35\text{ terms}\end{align}

These are triangle numbers, shunted by $1$, with the number of terms in $(x+y+z+w)^n$ calculated by $$\frac {(n+1)(n+2)(n+3)}{6}$$

Therefore, we have $(x+y+z+w)^{10}$ having $$\frac{11\times12\times13}6=286\text{ terms}$$