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Joined: 17 May 2017
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In how many different ways can a soccer team finish the season with
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Updated on: 13 Jun 2017, 05:43
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
(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.
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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Re: In how many different ways can a soccer team finish the season with
[#permalink]
13 Jun 2017, 06:43
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
(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:
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In how many different ways can a soccer team finish the season with
[#permalink]
13 Jun 2017, 05:07
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)
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)
