In how many different ways can a soccer team finish the

[swerve]

In how many different ways can a soccer team finish the season with three wins, two losses, and one draw?

A. 6
B. 20
C. 60
D. 120
E. 240

The OA is C

Source: Veritas Prep

[Scott@TargetTestPrep]

In how many different ways can a soccer team finish the season with three wins, two losses, and one draw?

A. 6
B. 20
C. 60
D. 120
E. 240

The OA is C

Source: Veritas Prep

We will use the permutation with indistinguishable items formula. Since there are 3 + 2 + 1 = 6 outcomes and the 3 wins and 2 losses cannot be distinguished, the total number of ways is given by 6!/(3!2!) = (6 x 5 x 4 x 3 x 2)/(3 x 2 x 2) = 5 x 4 x 3 = 60.

Answer: C

[GMATGuruNY]

In how many different ways can a soccer team finish the season with three wins, two losses, and one draw?

A. 6
B. 20
C. 60
D. 120
E. 240

Alternate approach:

There are 6 slots in the season: 3 wins, 2 losses and 1 draw.
From the 6 slots, the number of ways to choose 3 to be occupied by the 3 wins = 6C3 = (6*5*4)/(3*2*1) = 20.
From the 3 remaining slots, the number of ways to choose 2 to be occupied by the 2 losses = 3C2 = (3*2)/(2*1) = 3.
Since only 1 slot remains, the number of options for the draw = 1.
To combine the options above, we multiply:
20*3*1 = 60.

The correct answer is C.

[Brent@GMATPrepNow]

In how many different ways can a soccer team finish the season with three wins, two losses, and one draw?

A. 6
B. 20
C. 60
D. 120
E. 240

Question rephrased: In how many different ways can we arrange the letters WWWLLD

-------------ASIDE--------------------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in TOTAL
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

-------------NOW ONTO THE QUESTION!!--------------------------------------

WWWLLD
There are 6 letters in TOTAL
There are 3 identical W's
There are 2 identical L's
So, the total number of possible arrangements = 6!/[(3!)(2!)
= 60

Answer: C