Solving a Combinations Problem: 15 or 21?

[christian0710]

Homework Statement

Hi So the problem in combinations is the following: In how many ways can you choose 2 letters from the following 6 letters (A,B,C,D,E,F)

This is a question in combinations so I know the order does not matter, so AB is the same as BA.
I use the equation (n,r) = n!/((n-r)!*r!) = 15 (See my drawing)

The problem is this: If i draw out the possible combinations and add them together i get 21!

What is the correct answer 15 or 21, and why?

 

[ehild]

You can not choose the same letter twice or can you?

ehild

 

[christian0710]

Interesting. Does that mean that in combinations mathematics, you can never use the same 2 letters? Or at least acording the the formulae i use? Ahh so the logic is that you don't have 2 of the same letters in a group of letters, or course :)
Thank you for the hint.

 

[phinds]

Interesting. Does that mean that in combinations mathematics, you can never use the same 2 letters? Or at least acording the the formulae i use? Ahh so the logic is that you don't have 2 of the same letters in a group of letters, or course :)
Thank you for the hint.

It means you have to pay attention to the problem statement. Some problems using letters will allow letters to be used more than once, some will not. Problems that draw marbles out of a bag will generally imply that you can't put a marble back after you've used it. And so forth. The problem statement should be clear but in the absence of such clarity it's best to assume things can only be used once. With marbles, that's pretty clear, with letters not so much.

 

[ehild]

If you can choose the same letter again, it is "combination with repetition". The number of possibilities are equal to (n+r-1)/(r! (n-1-r)!). If n=6 and r=2, it is 21.

ehild