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?


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.

  • $\begingroup$ So that means greatest coefficient is $\frac{25!}{6! 6!6! 7!}$? $\endgroup$ – Mathematics Feb 28 at 16:45
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.