A cool "Number of terms" problem!

Guys there is a better solution, every term of this expansion can be written as $\frac { 10! }{ a!b!c!d! } { w }^{ a }{ x }^{ b }{ y }^{ c }{ z }^{ d }\\ such\quad that,\quad a+b+c+d\quad =\quad 10$ Here we are not bother about the coefficient, so the number of terms is the solution of the condition given below. No of non-negative solution of the equation $a+b+c+d = 10$ is $\left( \begin{matrix} 10-4+1 \\ 4-1 \end{matrix} \right) =286$

Here we divide $(w+x+y+z)$ into $(w+x)+(y+z)$ ,now we take the $n^{th}$ term of the binomial expansion(without the co-efficient),(since now there are only two terms),: $(w+x)^{n}\times(y+z)^{10-n}$ ,here we have neglected the co-efficient as it will have no effect on the number of terms. $$ Now,we notice that the first part i.e $(w+x)^n$ has $n+1$ terms and the latter part has $10-n+1=11-n$ terms.Hence the product would have $(n+1)(11-n)$ terms,hence the $n^{th}$ term has $-n^2+10n+11$ terms.Now,we just have to add this expression with $n$ ranging from $0\rightarrow 10$ i.e. $\sum_{n=0}^{10}-n^2+10n+11\\ =-\left(\dfrac{10\times11\times21}{6}+10\left(\dfrac{10\times11}{2}\right)+11\times11\right)\\ =286$ .And done!