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Re: In how many different ways can a soccer team finish the season with
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13 Jun 2017, 05:40
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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
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.
Re: In how many different ways can a soccer team finish the season with
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13 Jun 2017, 06:43
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Top Contributor
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
Re: In how many different ways can a soccer team finish the season with
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15 Jun 2017, 16:10
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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
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.
Re: In how many different ways can a soccer team finish the season with
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07 Dec 2018, 13:00
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