# Maximum value of coefficient in Multinomial Expansion

Find the maximum value of coefficient in the expansion of $$(x+y+z+w)^{25}$$.

Basically what the question is saying is that all term will be of type $$k x^{r_1}y^{r_2}z^{r_3}w^{r_4}$$ so what can be maximum value of $$k$$.

Well in binomial expansion, middle binomial coefficients are greatest but how to expand that thought here?

I wrote it as $$(a+b)^{25}$$, in this expansion the term having greatest coefficient will be $$C(25,13) (x+y)^{12}(z+w)^{13}$$ and then take maximum binomial coefficient of $$(x+y)^{12}(z+w)^{13}$$ to get answer as $$C(25,13) \times C(12,6) \times C(13,7)$$ but I am not sure it is correct. Could someone help me with this?

## 1 Answer

Your observation that in binomial coefficients, the central ones are largest is the key. Let's conjecture that the same holds here: the multinomial coefficient will be largest when the difference between any two of the $$r_i$$ is at most $$1$$. To prove this, suppose for example $$r_1-r_2\geq2.$$ Show that you get a larger coefficient if you replace $$r_1$$ by $$r_1+1$$ and $$r_2$$ by $$r_2-1,$$ leaving $$r_3,r_4$$ unchanged. This follows at once from your observation about the binomial coefficients.

• So that means greatest coefficient is $\frac{25!}{6! 6!6! 7!}$? – Mathematics Feb 28 at 16:45
• @Mathematics Yes, that's right. In general if we have $(x_1+x_2+\cdots+x_k)^n,$ write $n=qk+r, 0\leq r<k$. Then we we get the largest multinomial coefficients when $r$ of the $r_i$ equal $q+1$ and the other $r_i$ all equal $q.$ – saulspatz Feb 28 at 16:52