Multinomial Theorem and Coefficients of Polynomial Expansions
Date: 06/24/2007 at 19:23:36 From: Siddhant Subject: Multinomial Theorem Could you explain the Multinomial Theorem to me? For example, find (x + y + z + a + b)^4? I looked up the Multinomial Theorem, but can't quite understand it.
Date: 06/25/2007 at 02:12:12 From: Doctor Greenie Subject: Re: Multinomial Theorem Hello, Siddhant -- I found one page in the Dr. Math archives where this topic is very briefly discussed: Multinomial Coefficients http://mathforum.org/library/drmath/view/64616.html Let me go into much greater detail for you on your question.... Let's start with the binomial theorem. I assume if you are asking questions about the multinomial theorem that you are familiar with the binomial theorem. The binomial theorem says that the coefficient of the (x^m) * (y^(n-m)) term in the expansion of (x+y)^n is "n choose m", denoted variously as nCm or C(n,m) or as "n" above "m" in a single large set of parentheses. (I can't type that one.) The number nCm has the value n! ---------- (m!)(n-m)! By a well-known principle of combinatorics, this same expression is the number of distinct ways of arranging "n" objects of which "m" are identical and the remaining "n-m" are identical. Let's use this interpretation of this expression to see why the binomial coefficients are these nCm numbers. Suppose, for a specific example, we want to determine the coefficient of the x^2 * y^3 term in the expansion of (x+y)^5. Let's think of the expansion as the product of five factors of (x+y): (x+y)(x+y)(x+y)(x+y)(x+y) The product of any group of polynomials is the sum of the partial products formed by selecting one of the terms from each factor. In this example, we can pick either an x term or a y term from each of the five factors. If we want to count the number of partial products which are x^2 * y^3, then we need to select the "x" term from 2 of the 5 factors and the "y" term from 3 of the 5 factors. We could find the number of these partial products by using basic combinatoric principles; the number of these partial products is "5 choose 2", because we need to choose 2 of the 5 factors to be the ones from which we choose the "x" term. But using the "a choose b" concept will not help us when we move on to examine the multinomial coefficients. So let's count the number of partial products of the form x^2 * y^3 in a different way. We will write out all the ways we can select the "x" terms from 2 of the 5 factors. If we use the string of characters "xxyyy" to indicate selecting the x terms from the first two factors and the y terms from the last three factors, then the complete list of ways we can get a partial product of x^2 * y^3 is xxyyy xyxyy xyyxy xyyyx yxxyy yxyxy yxyyx yyxxy yyxyx yyyxx This is clearly the number of distinct arrangements of the letters "xxxyy", which we know is 5! -------- = 10 (2!)(3!) So the number of partial products of the form x^2 * y^3--and therefore the coefficient of the x^2 * y^3 term in the expansion of (x+y)^5--is the number of distinct ways of arranging 2 x's and 3 y's. The conclusion in the last paragraph makes the extension of our investigation to multinomial coefficients rather easy. Suppose we take your example to look at the question of multinomial coefficients: (x+y+z+a+b)^4 If we want to know the coefficient of the x^3 * a^1 term in this expansion, we know the number of partial products of that form is the number of distinct ways of arranging the letters "xxxa". And that number of ways is 4! -------- = 4 (3!)(1!) Of course, we don't need the "1!" in that computation; I simply showed it in this case for clarity. And similarly the coefficient of the x^1 * y^2 * b^1 in this expansion is 4! -- = 12 2! And the coefficient of the abxy term in the expansion is just 4! ---------------- = 24 (1!)(1!)(1!)(1!) Just a couple of random examples to make the method for determining multinomial coefficients more clear.... The coefficient of the a^2 * b^3 * c^4 term in the expansion of (a+b+c)^9 is the number of distinct arrangements of the letters "aabbbcccc", which is 9! ------------ = 1260 (2!)(3!)(4!) The coefficient of the a^4 * b^6 * c^2 * d^7 * e^3 term in the expansion of (a+b=c+d+e)^22 is the number of distinct arrangements of the letters "aaaabbbbbbccdddddddeee", which is 22! -------------------- = ??? (4!)(6!)(2!)(7!)(3!) I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
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