# Binomial Coefficients

1. Sep 21, 2005

### Gale

k, maybe wrong forum... whatever...

Anyway, so i was hoping someone could maybe derive or at least explain binomial coefficients. Like, i know that binomial(n,r)= n!/(n-r)!r! but why? in class the guy was explaining something like, if you're counting, and you're trying to arrange 3 balls into 2 groups then its like...

|1 2 3
1|2 3
1 2|3
1 2 3|

and that because of that, it was like, the formula we'd use for this problem would be binomial(n+r-1, r-1), and then we plug that into the factorial problem... and voila... but, i dunno. this didn't make sense to me at all. we introduce a divider, and we derive some formula for it... and... eh...

so, i see how the method was convienient, but i think there must be some more formal way of going about it. soo...

2. Sep 21, 2005

### hypermorphism

Okay, first we need to know that the set of all n-permutations of n objects (arrangements of all n objects) contains n! elements. It is easy enough to see this from the multiplication principle of counting. You can then derive the expression for all k-permutations of n objects where k is between n and 0 as n!/(n-k)!. This can be derived in an in situ fashion. Suppose x is the number of k-permutations. Then by the multiplication principle x*(n-k)! = n!, so we have x = n!/(n-k)!. (LHS explanation: x counts permutations of k things. Next, we permute n-k things for each permutation x counts. This is the same as the amount of permutations of n things.) This is the method we shall use to get the form for combinations.
Suppose you have a set of n objects. Let x be the number of k-combinations of n objects, where k is between n and 0. If we were to then count the k-permutations of these k objects, we should then have a number equivalent to k-permutations of the original n objects. Now each object in the set of combinations has k! permutations, so we have the equation k!*x = n!/(n-k)!, which we solve to get the familiar x = n!/[k!(n-k)!].

Last edited: Sep 21, 2005
3. Sep 26, 2005

### amcavoy

4. Sep 27, 2005

### robert Ihnot

You can look at it as a heads and tails situation. (H+T)(H+T) = HH+HT+TH+TT = H^2+2HT+T^2; which give us all of the four outcomes from flipping a series of heads and tails. This can all be repeated for the nth power, (H+T)^n. These are known, by the way, as Bernoulli trials.

5. Sep 27, 2005

### Neha Sanghvi

Uh, I am sorry, I am posting a question in this thread, but it's realted to binomial coeffecients. The question is that there exists an important result that (rootA + B)to the power n = I + f ( I - integral value, f - fractional part ). From this we get the value of (A + Bsquare)to the power n = K to the power n. Then, it is explained for n as odd and even integer separately.
I have not got this one clearly. Can someone plz explain this to me.