Multinomial CoefficientsDate: 02/02/2004 at 19:42:28 From: Igor Subject: Finding coefficients of polynomial terms What is the coefficient of the x^2*y^3*z^4 term in the (very poly!) polynomial (x+y+z)^9 ? I have figured out that if you keep the 3 y's and 4 z's at a certain location and just rearrange the x's there are 9C2 = 36 combinations possible and if you keep the x's and z's still there are 9C3 ways to move around the y's and same thing for the z's but where do I go from there? Please help: maybe there's some sort of theorem or something? Date: 02/02/2004 at 21:31:24 From: Doctor Vogler Subject: Re: Finding coeff. of term of (large polynomial) ^ large power Hi Igor, You seem to be familiar with the Binomial Theorem and binomial coefficients (called "n-choose-r"). What you need here is called "multinomial coefficients." Basically, if you have a sum of k variables (in your case, k=3), and you raise it to a power n (in your case, n=9), then you can calculate the coefficient on any particular term as follows: Notice that the sum of the exponents in any term will be n (as it is in your example, 2+3+4=9). Take the factorials of each of the exponents, multiply them together, and divide n! by the product. That is your coefficient. (Note how when k=2, you get the binomial coefficient.) For example, if you have three variables, (x + y + z)^n, then the coefficient of the x^i * y^j * z^k term (where i+j+k = n) is n! -------- i! j! k! Let me know if you'd like to talk about this some more, or if you have any other questions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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