Date: 02/25/2003 at 09:45:26 From: Sureyya Hikmet Subject: Trinomial expansion How can I find the number of terms of (a+b+c)^140 ?
Date: 02/26/2003 at 00:14:24 From: Doctor Greenie Subject: Re: Trinomial expansion Hello, Sureyya - Let's look at a much more simple example and then extend the analysis to your particular case. Let's find the number of terms in (a+b+c)^3. We will first list them all; then we will find a way to count the number of terms without listing them. The different terms in the expansion of (a+b+c)^3 are the following: a^3 b^3 c^3 (3) a^2b a^2c b^2a b^2c c^2a c^2b (6) abc (1) By listing them and counting, we see that there are 10 terms in the expansion of (a+b+c)^3. How do we find this number without listing the terms? We know that each term must contain 3 variable factors, each of which is either a, b, or c. In any term, if x = number of factors of a, y = number of factors of b, and z = number of factors of c, then we must have x+y+z = 3 So the number of terms in (a+b+c)^3 is equal to the number of ordered triples (x,y,z) of non-negative integers with x+y+z = 3. This in turn is equal to the number of ways of partitioning a set of 3 elements into 3 disjoint subsets. The number of ways of partitioning a set of 3 elements into 3 disjoint subsets can be modeled using 3 markers as the elements of the set and 2 markers as the 'dividers' between the subsets. If we use * for an element and | for a divider, then we have, for example, |**|* corresponds to the b^2c term (no elements in the 1st subset; two in the 2nd; one in the 3rd - i.e., no factors of a, two factors of b, and one factor of c) *||** corresponds to the ac^2 term (one element in the 1st subset; none in the 2nd; two in the 3rd - i.e., one factor of a, no factors of b, and two factors of c) So now we see that the number of terms in the expansion of (a+b+c)^3 is equal to the number of distinct arrangements of the characters ***|| This number of arrangements, as you probably know, is equal to 5! 120 -------- = ------ = 10 (3!)(2!) (6)(2) Having analyzed the relatively simple case of (a+b+c)^3, we can extend our reasoning to say that the number of terms of (a+b+c)^140 is equal to 142! (142)(141) ---------- = ---------- = (71)(141) = 10011 (140!)(2!) 2 I hope all 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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