Finding the number of different terms in multinomial expansion

Question

I am self studying Combinatorics from Richard Brualdi and I am unable to think about an argument in Chapter-5 .

In the section Multinomial theorem author writes -> Number of different terms that occur in Multinomial expansion of $ (x_1+ x_2 +... + x_t)^n $ equals the number of nonnegative integral solutions of $ n_1 +n_2 + ... + n_t = n. $

I am unanable to think how the number of distinct terms equal to the non negative solutions of given equation. Can somebody please explain!!

Answer 1

Like if ylu give $n_1$ value of 1 means the power of coefficient of $x_1$ is 1 , you give $n_2$ value 4 means power of $x_2$ in that term is 4. Each different case give different term .

here the $n_1,n_2,n_3....$ are whole numbers.

lets say if we add one to all numbers(making every solution natural) it effectively becomes $n_1+n_2+n_3.......n_t=n+t$( where $n_i$ is a natural number)

Now imagine there are n+t blanks and to put 't-1' vertical lines between them. (the number of blanks between $i-1^{th}$ line and $i^{th}$ is $n_i$ . But for $n_1$ its number of blanks before first line and $n_t$ is number of blanks left after last line.Obviously you cant fill first and last gap ). So number of solution is ${n+t-1 \choose t-1}$.

Answer 2

it's because of the following ...

$\underbrace{(x1+x_2+\cdots+x_t)(x1+x_2+\cdots+x_t)\cdots(x1+x_2+\cdots+x_t)(x1+x_2+\cdots+x_t)}_{\text {n times}}=(x1+x_2+\cdots+x_t)^n$

we pick 1 element from each parenthesized set, there are $n$ total so our exponents add up to $n$ that gets split over $t$ items just most are 0 and are never seen.