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There are 3 green flags, 2 red flags and 3 yellow flags. If all the f

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nick1816
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There are 3 green flags, 2 red flags and 3 yellow flags. If all the f [#permalink] New post   15 Jun 2019, 08:04
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There are 3 green flags, 2 red flags and 3 yellow flags. If all the flags are arranged in a row, what is probability that the flags at the ends are neither green nor yellow in color?

A. 1/60
B. 1/56
C. 1/30
D. 1/28
E. 1/15
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D
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Lampard42
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Re: There are 3 green flags, 2 red flags and 3 yellow flags. If all the f [#permalink] New post   15 Jun 2019, 08:43
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nick1816 wrote:
There are 3 green flags, 2 red flags and 3 yellow flags. If all the flags are arranged in a row, what is probability that the flags at the ends are neither green nor yellow in color?

A. 1/60
B. 1/56
C. 1/30
D. 1/28
E. 1/15


Total number of possibilities for arranging the flags = 8!/3!3!2! = 560
Possibility with restrictions as required= 6!/3!3!=20(red flag is arranged at ends)
So probability = 20/560=1/28 option D

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Re: There are 3 green flags, 2 red flags and 3 yellow flags. If all the f [#permalink] New post   15 Jun 2019, 08:47
Can you explain how you obtained possibility with restrictions?

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Re: There are 3 green flags, 2 red flags and 3 yellow flags. If all the f [#permalink] New post   15 Jun 2019, 08:50
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Shrey9 wrote:
Can you explain how you obtained possibility with restrictions?

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Red flags place is already determined that is one at each end. So balance left is 6 flags in sets of 3 identical items. So they can be arranged in 6!/3!3! Ways
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Re: There are 3 green flags, 2 red flags and 3 yellow flags. If all the f [#permalink] New post   16 Jun 2019, 15:28
The total number of ways to arrange 8 flags without repetitions is 8! or 8P8. However, it says the flags at the end cant be yellow or green,meaning they must be red. Hence,the flag at one extreme end can be chosen two ways,or 2P1 ways while the flag at the other extreme can be chosen in 1 way only. So we now have 2 flags at the extremes. We need to arrange the rest of the 6 flags which can be arranged in 6! ways or 6P6 ways. So the arrangement would be 6!*2 ways.
The probability would then be 6!*2/*8!= 1/28.
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Re: There are 3 green flags, 2 red flags and 3 yellow flags. If all the f [#permalink] New post   16 Jun 2019, 18:58
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The flags at each end need to be red. If we pick those flags first, there's a 2/8 chance the leftmost flag is red, and then a 1/7 chance the rightmost flag is red, so the answer is (2/8)(1/7) = 1//28. The other flags don't matter.
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Re: There are 3 green flags, 2 red flags and 3 yellow flags. If all the f [#permalink] New post   10 Jul 2020, 04:44
Hello from the GMAT Club BumpBot!

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Re: There are 3 green flags, 2 red flags and 3 yellow flags. If all the f [#permalink]
10 Jul 2020, 04:44
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