Download Finding binomial coefficients in (a+b)^n and more Study notes Mathematics in PDF only on Docsity!
Finding binomial coefficients in (a+b)n
Introduction-
Let’s see some binomial expansions like (a+b)^2 = a^2 + b^2 + 2ab & (a+b)^3 = a^3 + b^3 + 3ab^2 + 3ab^2. And focus on coefficient of ab^2 term in (a+b)^3 expansion, above: (a+b)^3 = (a+b).(a+b).(a+b). To get ab^2 from (a+b).(a+b).(a+b), we have three ways as: {(a, b, b), (b, a, b), (b, b, a)}.
In above, (a, b, b) from the first way; means we take ‘a’ from first (a, b), ‘b’ from second (a, b), & ‘b’ from 3rd (a, b) in order. It’s like selecting one blue ball(a) and 2 green balls(b) from three packets, where each packet contains 1 green & 1 blue ball. For each term(like ab^2) in (a+b)^3, we can see that we have to select 3 balls for each term.
One thing remains fixed: number of blue balls(a) + number of green balls(b) = n & 0<=number of blue balls(a)<=n & 0<=number of green balls(b)<=n. Here, n=3.
So, the coefficient of (i blue balls(a) + j green balls(b)) or a^i b^j is just counting the number of ways to select i a’s and j b’s from n packets, where each packet contains 1 a & 1 b & i+j=n & 0<=i,j<=n.
Counting the number of ways to select i a’s and j b’s from n
packets-
Returning to example: coefficient of ab^2 term in (a+b)^3 expansion?
To get ab^2 from (a+b).(a+b).(a+b), we have three ways as: {(a, b, b), (b, a, b), (b, b, a)} or counting the number of ways to select 1 a and 2 b’s from 3 packets, where each packet contains 1 a & 1 b.
If A, B, C were three elements of a set, then the number of ordered arrangements of (A, B, C) = 3! = 321 = 6. 3!, because for the first place in (_, _, _) arrangement, we have: 3 options then for 2nd we have: 2 and for 3rd we have only: 1 option. We multiply these counts/options to get the total number of arrangements. These 6 arrangements are as follows: {(A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), (C, B, A)}.
Similarly ordered arrangements of 1 a & 2 b's are {(a, b, b), (a, b, b), (b, a, b), (b, b, a), (b, a, b), (b, b, a)}. But some of these arrangements are repeated like (a, b, b), (b, a, b), (b, b, a). So, we must remove duplicate arrangements to get desired arrangements. For this we must find a way to prevent over counting!
Counting and removing duplicate arrangements:
Let’s say we pick (a, b, b) arrangement then number of ways in which 2 b’s can occur by switching places, keeping a’s position fixed = 2! Similarly for (b, a, b), (b, b, a) arrangements also. So now previous arrangements are like {(a, b, b)X2, (b, a, b)X2, (b, b, a)X2}, where X2 means twice counted. So, these 2! = 2 duplicate arrangements are the same for each of (a, b, b), (b, a, b), (b, b, a) arrangements.
