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In how many different ways can a soccer team finish the season with

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In how many different ways can a soccer team finish the season with  [#permalink]

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Updated on: 13 Jun 2017, 05:43
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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

Originally posted by haardiksharma on 13 Jun 2017, 04:58.
Last edited by Bunuel on 13 Jun 2017, 05:43, edited 2 times in total.
Renamed the topic, edited the question and moved to PS forum.
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In how many different ways can a soccer team finish the season with  [#permalink]

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13 Jun 2017, 05:07
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Let's call a win(W), a loss(L) and a draw(D)

The ways in which the team can finish the season with 3W,2L and 1D is $$\frac{{6!}}{{2!*3!}}$$ as every win or loss is alike.

Now, $$\frac{{6!}}{{2!*3!}}$$ = $$\frac{{6*5*4*3!}}{{2!*3!}}$$ = $$\frac{{6*5*4}}{{2}}$$ = 60(Option C)
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Re: In how many different ways can a soccer team finish the season with  [#permalink]

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13 Jun 2017, 05:40
haardiksharma wrote:
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 number of arrangements of Win, Win, Win, Loss, Loss, Draw, or the number of arrangements of 6 letters WWWLLD, where three W's and two L's are identical, is 6!/(3!2!) = 60.

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Re: In how many different ways can a soccer team finish the season with  [#permalink]

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13 Jun 2017, 06:43
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haardiksharma wrote:
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

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Re: In how many different ways can a soccer team finish the season with  [#permalink]

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15 Jun 2017, 16:10
1
haardiksharma wrote:
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

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.

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Re: In how many different ways can a soccer team finish the season with  [#permalink]

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07 Dec 2018, 13:00
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Re: In how many different ways can a soccer team finish the season with   [#permalink] 07 Dec 2018, 13:00
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