# Permutation and Combination Problem

1. Apr 27, 2012

### Mathkid182

1. The problem statement, all variables and given/known data

How many way can a team lose 3 of their next 5 games?
How many ways can a team lose 2 of their next 5 games?
Why are the two answers the same?

2. Relevant equations

Permutation = nPr = n! / (n-r)!
Combination = nCr = nPr / r!
where,
n, r are non negative integers and r<=n.
r is the size of each permutation.
n is the size of the set from which elements are permuted.
! is the factorial operator.

3. The attempt at a solution
I think this is a problem where the permutation equation would be used. After all, the team can LLLWW or WWLLL and those would be two different permutations that would satisfy the question's requirements. But I just don't understand how I would go about solving it. Any help is much appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 28, 2012

### Vorde

When I was studying for the SAT's and dealt with these equations the trick was to remember "n choose r", meaning that in the equations n is the number you are choosing from and r is the amount you are choosing.

In the first example you are choosing 3 out of 5 so n=5 r=3

Get it?

3. Apr 28, 2012

### Mathkid182

But then if we did the permutation formula for both questions then one wouldn't receive the same answer for both.

4. Apr 28, 2012

### HallsofIvy

Staff Emeritus
If you write "L" for lose and "W" for win you can see that the problem, is the same as 'In how many ways can you write 2 "W"s and 3 "L"s?'

If you imagine that all of the "L" are labeled, say "1", "2", "3", and all of the "W" are labeled "1", "2", then you have 5 distinguishable objects. There are 5! ways of arranging them. But because the two "W"s are not distinguishable, such an arrangement as $W_1L_1L_2L_3W_2$ is exactly the same as $W_2L_1L_2L_3W_1$. That is, because there are 2!= 2 ways to permute just the "W"s, we have to divide by 2! to get the number or arrangements ignoring rearrangements of just the "W"s. Similarly, there are 3!=6 ways to rearrange just the different "L"s. We must also divide by 3! since those are not "different" arrangements.

Now do the same with 2 "L"s and 3 "W"s.

5. Apr 28, 2012

### Villyer

If you were looking for all three-length permutations of {1,2,3,4,5}, you would get 123, 132, 213, 231, 312, 321 as six different possibilities.
Are all six of these different ways of winning three games?