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From the multinomial formula we have

$$(x_1 + x_2 + \dotsb + x_m)^n = \sum_{k_1+k_2+\dotsb+k_m=n, \ k_1, k_2, \dotsc, k_m \geq 0} {n \choose k_1, k_2, \dotsc, k_m} \prod_{t=1}^m x_t^{k_t}\,.$$

I wish to evaluate the RHS, but with the additional constraint that each exponent $k_i$ can not exceed some positive integer $a<n$. That is, $$ \sum_{k_1+k_2+\dotsb+k_m=n, \ 0\leq k_1, k_2, \dotsc, k_m \leq a} {n \choose k_1, k_2, \dotsc, k_m} \prod_{t=1}^m x_t^{k_t}\,. $$ While I can obviously calculate this, my question is whether there is a simple way to express the polynomial with the restriction on the exponents. This is to ultimately facilitate computation. For example, in the case where $a=n-1$ we could write this as $$ \left(\sum_{i=1}^m x_i\right)^n - \sum_{i=1}^m x_i^n. $$ Is there a simple way of expressing the polynomial for other integer values $0< a<n$?

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    $\begingroup$ Even for the binomial case, you can't really simplify $\sum_{n-a\leq k \leq a} \binom{n}{k} x^k$ in any meaningful way. $\endgroup$ Commented Mar 15, 2022 at 22:14

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The idea of subtracting large powers from the multinomial can be extended with the inclusion-exclusion principle:

$$ \sum_{k_1+k_2+\dotsb+k_m=n,\atop 0\leq k_1, k_2, \dotsc, k_m \leq a} {n \choose k_1, k_2, \dotsc, k_m} \prod_{t=1}^m x_t^{k_t} = n! \sum_{S\subseteq [k]} (-1)^{|S|} \sum_{t_s\geq a+1, s\in S\atop \sum_{s\in S} t_s\leq n} \frac{\big(\sum_{i=1\atop i\not\in S}^m x_i\big)^{n-\sum_{s\in S} t_s}}{(n-\sum_{s\in S}t_s)!} \prod_{s\in S} \frac{x_s^{t_s}}{t_s!}. $$

If we introduce a truncated exponential series $\exp_l(x) := \sum_{t\geq l} \frac{x^t}{t!}$. Then the above formula can be seen as $$[z^n]\ n!\sum_{S\subseteq [k]} (-1)^{|S|}\exp\big(z\sum_{i=1\atop i\not\in S}^m x_i\big) \prod_{s\in S} \exp_{a+1}(x_sz).$$

I'm not sure how helpful it is.

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