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?

1. 1/60
2. 1/56
3. 1/30
4. 1/28
5. 1/15

Correct Answer: D
Solution and Explanation:
Approach Solution 1:
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

Approach Solution 2:
The total number of ways to arrange 8 flags without repetitions is 8! or 8P8. However, it says the flags at the end can't 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.

Approach Solution 3:
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

“There are 3 green flags, 2 red flags and 3 yellow flags. If all the”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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