Don't expand it, please!

Use the Multinomial theorem

Number of terms in $(x_1+x_2+....+x_m)^n= \binom{n+m-1}{n}$

In the given ques., $m=4,n=10$

$\implies$ No. of terms $=\binom{10+4-1}{10}=\dfrac{13!}{10!.3!}=\boxed{286}$

Well we can simply see that every terms is in the form $x^ay^bz^ct^d$ , where $a+b+c+d=10$ . So we simply find the number of non-negative $(a,b,c,d)$ satisfying this. This is simply: $13C10=286$

We can conclude x+y+z+t as a+b+c+d= 10. If the whole square is also 10. then we can conclude abcd as non negative no ( a,b,c,d) then we can say 13c10=286.

suggestion-question must be asked as "distinct terms"

Using stars and bars

(1,2,7,0)

These numbers represent the stars and after each number of stars, we have a bar. So for this example, we have:

let o represent a star

o | o o | o o o o o o o |

We have 10 stars and 3 bars.

Stars + bars C Stars = 10+3 C 10

13C10 = 286

use begger method