3where n_{1}+n_{2}+.....n_{m}=n
n1,n2...nm are all non negative integers.
If ni is a positive integer and a1,a2....am belongs to C. then
(a1+a2+.......+an)n=\Sigma n! a_{1}^{n_{1}}a_{2}^{n_{2}}....a_{m}^{n_{m}}/(n_{1}!.n_{2}!...n_{m}!).
pls give the combinatorial as well as mathematical proof for the eqn.
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3 Answers
1luk,
for any set of powers ni
where sum(ni)=n
say we have 'n' brackets of the type (a1 +a2+a3..) multiplied amongst them selves..to give (a1 +a2+a3..)n
we have
to pick out n1 brackets out of the n brackets that will contribute n1 "a1's "...
to do that we have nCn1 ways of picking, now we are left wid, n-n1 other brackets out of which we have to pick up n2 other brackets that will give us
n2 number of "a2 's"....
can be done in n-n1Cn2
again we are left wid n-(n1+n2) brackets out of which we need to select n3 brackets dat will give us "n3" number of "a3 's"
can be done again in
n-(n1+n2)Cn3 ways...
so the final coefficient of
a1n1*a2n2*....aknk
=
nCn1*n-n1Cn2*n-(n1+n2)Cn3*....
ways..
which comes out to be
the expression given above in the question..
so dis is the combinatorial proof...
cheers!!
3for gordo actually i cant understand the combinatorial interpretation for the multinomial theorem in both the qns. so pls give an idea abt the combinational interpretation.