In how many solutions of equation : $x_1+x_2+...+x_n=m$

Question

In how many solutions of equation: $x_1+x_2+...+x_n=m$ satisfied: $x_i\in \mathbb{N}(i=\overline{1,n}),1\le x_i\le 26,n\le m\le 26n,m\in \mathbb{N}$

This is my try: Let $t_i=x_i-1\implies \sum\limits_{i=1}^{n}t_i=\sum\limits_{i=1}^{n}x_i-n=m-n$, where $0\le t_i\le 25$. And I don't know how to solve it when $0\le t_i\le 25$.

Can you help me how to compute coefficient of $x^{m}$ in $(1+x+...+x^{26})^n$ ? — know dont, Apr 13 '19 at 23:02
There is no easy way to do that. You can take the derivative $m$ times for example, but that is still just as hard. — Peter Foreman, Apr 13 '19 at 23:19
I'm really grateful for all comments, I have just found it :https://math.stackexchange.com/questions/1868044/the-number-of-n-tuples-of-nonnegative-integers-x-1-ldots-x-n-satisfying-t?rq=1 — know dont, Apr 13 '19 at 23:20

Joel Pereira · Answer 1 · 2019-04-14T14:48:35.333

2

The question is asking how many integer solutions are there for $\displaystyle\sum_{i=1}^nx_i$ = m. We can use combinations with repetition. To find the number of such solutions, we can think of $m+n-1$ slots and we pick $n-1$ "dividers". The number of slots between the dividers d$_i$ and d$_{i+1}$ = x$_{i+1}$. We designate the slots before the first divider as x$_1$ and after the last divider as x$_n$. Thus every arrangements of lots and dividers is a different solution. So the number of solutions = $\binom{n+m-1}{n-1}$ = $\binom{n+m-1}{m}$.

edited Apr 14 '19 at 14:48

answered Apr 13 '19 at 23:23

Joel Pereira

1,229

This doesn't use OP restriction $x_i \le 26.$ – coffeemath Apr 14 '19 at 02:05
1

it's clear that if $m \le$26, then the $x_i \le$ 26. – Joel Pereira Apr 14 '19 at 04:03
Joel-- yes, just noticed that, will delete comment. Meanwhile +1. – coffeemath Apr 14 '19 at 06:37

In how many solutions of equation : $x_1+x_2+...+x_n=m$

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