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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?

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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.

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  • $\begingroup$ So that means greatest coefficient is $\frac{25!}{6! 6!6! 7!}$? $\endgroup$ – Mathematics Feb 28 at 16:45
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    $\begingroup$ @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 $q.$ $\endgroup$ – saulspatz Feb 28 at 16:52

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