Coefficients in a Trinomial Expansion
Date: 04/24/2001 at 14:34:48 From: Mark Birkenfeld Subject: Coefficient in trinomial expansion In the expansion of (a+b+c)^6, what is the coefficient of a^2.b^2.c^2? Without taking 10 or 15 minutes to multiply this out on paper, how can I find the answer? I figure that the answer is related to the binomial theorem and Pascal's triangle, but I cannot find any text to back up my hypothesis.
Date: 04/24/2001 at 16:44:44 From: Doctor Rick Subject: Re: Coefficient in trinomial expansion Hi, Mark. I would solve the problem using combinatorics. The question is equivalent to this one: How many different six-letter "words" can be formed using two a's, two b's, and two c's? To answer it, first pretend the two a's are different (a1 and a2), and similarly for the b's and the c's. Then there are 6! permutations of the six distinct symbols. But any one of these "words" is in fact identical to the "word" obtained by swapping the two a's; thus we must divide the total by 2. Likewise we must divide by 2 again to account for the identical b's, and again to account for the identical c's. The final answer is: 6!/2^3 = 6*5*4*3*2/(2*2*2) = 90 You can see the connection to the binomial theorem: for example, the coefficient of a^3.b^3 in the expansion of (a+b)^6 is: /6\ 6! ( ) = ----- = 6*5*4*3*2/(3*2*3*2) = 20 \3/ 3! 3! Here we again start with the permutations of six symbols. This time we divide by the number of ways that the three a's can be permuted, and the number of ways that the three b's can be permuted. We're already very close to coming up with a general formula for the coefficient of any term in the trinomial expansion. We just don't have a single symbol for the answer, equivalent to the "combinations" symbol. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 04/24/2001 at 16:46:57 From: Doctor Greenie Subject: Re: Coefficient in trinomial expansion Hi, Mark -- Yes, you are right - this is very closely related to the binomial theorem, Pascal's triangle, and "m choose n." Let's look first at the more familiar binomial theorem, Pascal's triangle, and the expansion of (a+b)^n. As you know, the expansion of (a+b)^n is: (a+b)^n = C(n,n)*a^n + C(n,n-1)*a^(n-1)b + ... + C(n,0)*b^n The coefficient of the a^(n-k).b^k term in this expansion is n! C(n,n-k) = --------- (n-k)!*k! One place where this number C(n,n-k) comes up is as the number of distinct ways to arrange (n-k) objects of one type and (k) objects of another type - for example, the number of distinct arrangements of the letters "aaaaabbb" is C(8,3) or C(8,5). Let's try to draw a parallel between the expansion of (a+b)^n and the arranging of a collection of two different kinds of objects. Let's think of (a+b)^5 written out as the product of five binomials: (a+b)^5 = (a+b)*(a+b)*(a+b)*(a+b)*(a+b) Now let's think of looking at this multiplication and see if we can determine the coefficient of the a^3.b^2 term in the expansion. As an extension of the familiar FOIL method for multiplying two binomials, the expansion of (a+b)^5 is found by summing all the partial products formed by choosing one of the two terms in each of the five binomial factors. So in expanding (a+b)^5, we can write down all the partial products that will contain three factors of a and two factors of b with the following notation: ababa will be the partial product formed by taking factors of a from the first, third, and fifth binomials and factors of b from the second and fourth. Then our list of partial products containing three factors of a and two factors of b is the following: aaabb aabab aabba abaab ababa abbaa baaab baaba babaa bbaaa There are ten such partial products. 10 is C(5,3); specifically, 10 is the number of distinct arrangements of 3 a's and 2 b's. With our notation for partial products as in the preceding list, there is a one-to-one correspondence between the partial products a^3.b^2 and the arrangements of three a's and two b's. So we have: coefficient of (a^3b^2) term in (a+b)^5 = # partial products with 3 factors of 'a' and 2 factors of 'b' = # arrangements of letters "aaabb" 5! = ---- 3!2! = 10 Now we can expand this idea to the expansion of (a+b+c)^n, or (a+b+c+d+e+f)^n, or to the expansion of any polynomial. For your particular problem, you have (a+b+c)^6, which you can write as: (a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c) If you think of using the extension of the FOIL method with this multiplication, and if you want to find the coefficient of the a^2.b^2.c^2 term, then you want to count the number of ways you can form a partial product by choosing the a from two of the six trinomial factors, the b from two of the other six, and the c from the remaining two factors. But this number of ways is simply the number of distinct arrangements possible of the letters aabbcc. Here are a few examples of the general application of this idea (you are already familiar with the first example, because of the binomial theorem): in the expansion the coefficient of of... the... term is... ----------------- ------------------ -------------------- 5! (a+b)^5 a^3.b^2 ---- = 10 3!2! 6! (a+b+c)^6 a^4.bc ------ = 30 4!1!1! 10! (a+b+c+d+e+f)^10 a.b^4.d^2.f^3 ------------ = ? 1!4!0!2!0!3! (Note the 0! factors in the denominator of the last example, indicating 0 factors of c and 0 factors of e. This is one example of why 0! is defined to have the value 1.) I hope this helps. Write back if you have more questions. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum