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

Question

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

Lampard42 · Answer

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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Shrey9 · Answer

Can you explain how you obtained possibility with restrictions?

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Lampard42 · Answer

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

Shishou · Answer

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.

IanStewart · Answer

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.

PSKhore · Answer

Total possibilities for picking flags at the end would be 3+2+3 = 8
8C2 = 28

Out of this one possibility would be picking both the red flags

=1/28

Alexdb10 · Answer

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

My working out was right up until the last part.

I recognise that there are 8!/3!3!2! ways to arrange the flags

then I recognised that if a red flag were to be at each end, there would be 6!/3!3! ways to do so.

However, surely it would be 2*6!/3!3! ways? as the red flags can be switched round?

Bunuel · Answer

You don’t multiply by two because the red-at-start and red-at-end arrangement is identical either way. “Red 1 at the start and Red 2 at the end” is the same arrangement as “Red 2 at the start and Red 1 at the end.”

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

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

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