Multinomial expansion

**alexmahone** · August 19th 2011, 03:56 PM

If we expand the expression

$(x_1+x_2+x_3+x_4)^6$

what will be the largest coefficient that occurs?

My attempt:

My hunch is that the coefficient should belong to one of the middle terms.

$\dbinom{6}{2,2,1,1}=\frac{6!}{2!2!1!1!}=180$

Is it right? I'm not sure how to prove it, though.

**Prove It** · August 19th 2011, 06:32 PM

Originally Posted by alexmahone

If we expand the expression

$(x_1+x_2+x_3+x_4)^6$

what will be the largest coefficient that occurs?

My attempt:

My hunch is that the coefficient should belong to one of the middle terms.

$\dbinom{6}{2,2,1,1}=\frac{6!}{2!2!1!1!}=180$

Is it right? I'm not sure how to prove it, though.

Expanding this would be a pain. I would write it as $\displaystyle (X_1 + X_2)^6$ , with $\displaystyle X_1 = x_1+ x_2$ and $\displaystyle X_2 = x_3 + x_4$ . Then you can use a binomial expansion, and when you back-substitute, you'll end up with more binomials you can expand.

**alexmahone** · August 19th 2011, 06:39 PM

Originally Posted by Prove It

Expanding this would be a pain. I would write it as $\displaystyle (X_1 + X_2)^6$ , with $\displaystyle X_1 = x_1+ x_2$ and $\displaystyle X_2 = x_3 + x_4$ . Then you can use a binomial expansion, and when you back-substitute, you'll end up with more binomials you can expand.

The question only asks for the largest coefficient, not for the expansion.

**Prove It** · August 19th 2011, 06:45 PM

Originally Posted by alexmahone

The question only asks for the largest coefficient, not for the expansion.

And yet, expanding is the only way I can think of to get the largest coefficient...

**Plato** · August 20th 2011, 02:32 AM

Originally Posted by alexmahone

If we expand the expression
$(x_1+x_2+x_3+x_4)^6$
what will be the largest coefficient that occurs?
$\dbinom{6}{2,2,1,1}=\frac{6!}{2!2!1!1!}=180$
Is it right?

Yes, that is correct. Note that is the 'smallest' denominator possible.

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