Evaluating 12 C 2: How to Simplify Without a Calculator

[qwexor]

Ok just a basic question. We are given 12 C 2 to evaluate. Which equals

12! / (10!2!)

Is there an easier way to simplify it. We are asked to do it without a calculator.

Thanks.

 

[cyby]

Compute it. Write it all out. Lots of stuff to cancel out.

You'll end up with 12*11/2 = 6*11 = 66

 

[HallsofIvy]

Cancel everything you can with the larger factorial in the denominator.

12!/10!= 12*11 so 12!/(10!)(2!)= 12*11/2= 6*11= 66.

 

[Ethereal]

This is something of an elementary question, but why is it always true that nCr where n>r would always yield a positive integer?

 

[shmoe]

That's a good question. The identity (n+1)Cr=nCr+nC(r-1), valid if n>=1 and 1<=r<=n, is easy enough to prove using nCr=n!/((n-r)! r!). Also, nC0=nCn=1 for all n>=0. These facts embody the construction of Pascal's triangle and let you prove by induction (on n) that nCr is always an integer since it's either 1 or it's the sum of two integers.

You can also think of the combinatorial definition of nCr, the number of r-subsets of an n-set. This is definitely an integer. Of course then you'd have to prove that the factorial expression for nCr is correct...

 

[Tide]

This is something of an elementary question, but why is it always true that nCr where n>r would always yield a positive integer?

The nCr are binomial coefficients with the property that n+1Cr = nCr-1 + nCr (think of Pascal's Triangle) with nCn = 1 = nC0. Since 0C0 = 1 it follows that all nCr are integers.

 

[Tide]

Shmoe beat me to it! :-)