combination or permutation

Question

How many groups of three pens can you select from a box of 12 pens? Find each answer 

Answer 1

Hmmm, first, do you want combinations (the order in the group of 3 doesn't matter), or permutations (group order matters). In general the number of combinations is less than the number of permutations for a given group size.

 

Let's do combinations first. We're calculating C(12,3) or the number of combinations taken 3 at a time from a set of 12.

The general formula is C(n,x) = n!/((n-x)!*x!)

 

C(12,3) = 12! /( (12-3)! 3!) = 12! / (9!)(3!) where ! indicates factorial. Notice that the 9! in the denominator will cancel out the first 9 factors in the numerator so we'll be left with (10*11*12)/(1*2*3) = (5*11*4)/1 = 220.

 

For permutations, P(n,x) = n!/(n-x)!, so P(12,3) = 12!/(12-3)! = 12!/9!. The factors 1-9 will cancel leaving 10*11*12 = 1320 permutations.

Answer 2

This is a counting problem. In this case, the order of the pens does not matter.

That is the important part: whether or not the order matters.

 

For example: what if we pick out a red, blue, and black pen?

 

What if we pick the red pen first, the blue pen second, and the black pen third?

Is that the same as the blue pen first, the black pen second, and the red pen third?

If they count as different results, then it is a permutation: the order we select matters.

If they count as the same result, then it is a combination: the order we select doesn't matter.

 

The problem asks for how many groups of pens you can select, it doesn't matter which order we select them from the box. So, this is a combination.

 

So, since we are choosing 3 pens from 12 possible pens, the answer is going to be "12 choose 3", which we write as ₁₂C₃

You can plug this into a calculator or use the formula for combination we get, ₁₂C₃=(12!)/((12-3)!3!)=220 as our answer.