multinomial?

**skystar** · May 31st 2008, 08:59 AM

$F:=F(A,B,C,D)=(A+B+C+D)^{6}$ and $G:=G(A,B,C,D)=(A^{2}+B+C^{2}+D)^{4}$

determine the coefficients with which the following terms appear in F and G.
i) $A^{2}BC^{2}D$
ii) $A^{2}B^{2}CD$
iii) $A^{2}B^{3}$

i have some lecture notes on this but would find it great help if someone could show me how,also so i could double check my answers when i get them.

**Plato** · May 31st 2008, 09:27 AM

Any term in the expansion of $\left( {A + B + C + D} \right)^6$ appears as $\frac{{6!}}{{g!h!j!k!}}A^g B^h C^j D^k$ where the non-negative integers $g + h + j + k = 6$ .

**skystar** · May 31st 2008, 09:32 AM

but what about an example for G when there is a power included?

**Plato** · May 31st 2008, 09:36 AM

Originally Posted by skystar

but what about an example for G when there is a power included?

How about your telling us what you think it might mean.

**skystar** · May 31st 2008, 09:39 AM

well for iii) and using G,

then 4!/(4!)(4!)

would the powers get added to one another.

**Plato** · May 31st 2008, 09:50 AM

Any term in the expansion of $\left( {A^2 + B + C^2 + D} \right)^4$ appears as $\frac{{4!}}{{g!h!j!k!}}A^{2g} B^h C^{2j} D^k$ where the non-negative integers $g + h + j + k = 4$ .

ii) cannot appear in g! Do you see why?
iii) cannot appear in f, but it can in g! Do you see why?

**skystar** · May 31st 2008, 10:29 AM

im still getting really confused as to why it cant appear. for iii in G

i do (4!/2!3!) ((A^4)(B^3)(C^2)(D))

$=2 A^{4}B^{3}C^{2}D$
thanks

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