proof needd

luk,
for any set of powers n_i
where sum(n_i)=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 ⁿC_n1 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-n1C_n2

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)C_n3 ways...
so the final coefficient of
a1ⁿ¹*a2ⁿ²*....ak^nk
=
ⁿC_n1*^n-n1C_n2*^n-(n1+n2)C_n3*....
ways..
which comes out to be
the expression given above in the question..
so dis is the combinatorial proof...
cheers!!

for 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.