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anonymous
 5 years ago
how can i deal with the binomial theorem if there are three or more variables
anonymous
 5 years ago
how can i deal with the binomial theorem if there are three or more variables

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you need help with?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(x+2y+3z+t^2+2)^32 how many terms in the expansion of the polynomial have ypart exactly y^8 (such as x^24y^8, x^12y^8z^8 , ...)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ARe you looking for the coefficients for the y^8 terms? Or how many terms have a y^8 term?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i wanna know how many terms it has. how can i solve such a problem like this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there would be four terms with y^8 terms, because , i think, binomial theorem states every variable there is one descending varible per term until it is vanquiched.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the number of variables in the problem,in your case four, so there are four y^8 terms. I am not 100% sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ (x+y)^4 = x^4+ 4 x^3y + 6 x^2 y^2 + 4 x y^3 + y^4. \] the x terms are decreasing until they are done this would happen to all the z, x, and t with y^8 being contant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i mind you y^8 an occur too!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this was the question that was asked last year in my department and i found the solution but i cant get it. let me write the solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the terms are x^n.(2y)^8.(3z)^m.(t^2)^k.2^p n+m+k+p+8=32 n,m,k,p in Z and greater than or equal to 0 n+m+k+p=24 number of solutions C(4+241, 24) Thus number of terms C(27,24)
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