Answer Probability Quiz: ((x1 + x2 +... + x33)^4) Monomials & Sum

[ParisSpart]

In the expansion of ((x1 + x2 +... + x33) ^ 4) how many monomials are of the form (xi ^ 2)*(xj^2)
with i not equal with j

if we add the coefficients of all these mononymon what is the sum?

E.g. the expansion of (x1 + x2 + x3) ^ 4 the requested number is 18

this is a probability quiz but i can think how to solve it with theory of probabilities any ideas for howmto beggin to solve this?

 

[mfb]

There is nothing probabilistic here, but combinatorics is important.
Have a closer look at your example - you should see some pattern about the coefficients. Find that pattern, and extend it to 33.

 

[haruspex]

E.g. the expansion of (x1 + x2 + x3) ^ 4 the requested number is 18

I don't think so. I believe 6x₁²x₂² constitutes one monomial.

 

[Bacle2]

I think it comes down to counting the number of ways of selecting a pair from {1,2,...,33}. For each

pair (x<sub>i[/SUBi,xj)) ,there will be one term xi[/SUP]2[/SUP]*xj² . The general binomial coefficient ci in

ci x1^y1x2^y2x3^y3 x4^y4 counts

precisely the number of ways of selecting a total of y_1 x_1's, y_2 x_2's, etc. in the product.

Then ,for

xi²xj², you're counting the number of ways of selecting exactly 2 x_i's and 2 x_j's from the expansion :

(x1+x2+x3+x4)... ( 4 times )

How many ways can you choose a pair (xi,xj) from

(x1,x2, x3,...,x33)?</sub>