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06 May 2024, 09:31
Kudos
Bookmarks
_ _ _ _ _ _ (6 places in a queue)
J should be ahead of F
Let's select any 2 spaces out of 6 for J and F: 6C2
The way to arrange these 2 people: 1 (Since J should always be ahead of F.)
Then for remaning 4 places, 4 other people can be arranged in 4! ways
Therefore, answer is (Selection of 2 places out of 6) * (Their arrangement) * (Arrangement of other 4 people)
= 6C2 * 1 * 4!
= (6*5/2) * 1 * (4*3*2)
= 360
(Once you understand the process, its easy to solve it in two lines. I would directly make 6 places (for picturising - not required), write 6C2*1*4! and then answer.)
Hope it helps!
J should be ahead of F
Let's select any 2 spaces out of 6 for J and F: 6C2
The way to arrange these 2 people: 1 (Since J should always be ahead of F.)
Then for remaning 4 places, 4 other people can be arranged in 4! ways
Therefore, answer is (Selection of 2 places out of 6) * (Their arrangement) * (Arrangement of other 4 people)
= 6C2 * 1 * 4!
= (6*5/2) * 1 * (4*3*2)
= 360
(Once you understand the process, its easy to solve it in two lines. I would directly make 6 places (for picturising - not required), write 6C2*1*4! and then answer.)
Hope it helps!
27 Apr 2025, 10:34
Kudos
Bookmarks
benbitz
You need to train yourself to recognize symmetry in problems like this.
When there are two people and no specific position restrictions (like “must be directly behind”), then in any full arrangement, either one will be ahead of the other, equally likely.
It’s not about detailed counting, it’s about realizing that between any two distinct people, the probability of one being ahead of the other is exactly 1/2.
With practice, you’ll start spotting this pattern more easily.
