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Bunuel
The 5 letters in the list G, H, I, J, K are to be rearranged so that G is the 3rd letter in the list and H is not next to G. How many such rearrangements are

A. 60
B. 36
C. 24
D. 12
E. 6
We have two case here...when h is at first place and when h is at second place
H at first place=6 ways
H at second place = 6 ways
Total= 12 ways
So...ans is D

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We have 5 letters: G, H, I J, and K.

Condition 1: G at \(3^{rd}\) place. ( There is only one way we can place G = 1!)
Condition 2: H is not next to G ( H can be placed at \(1^{st}\) or \(5^{th}\) position = 2!)

Remaining three letters I, J and K can be arranged in 3! ways.

Overall ways: 1! *2! *3! = 12 ways.

Answer D
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Bunuel
The 5 letters in the list G, H, I, J, K are to be rearranged so that G is the 3rd letter in the list and H is not next to G. How many such rearrangements are

A. 60
B. 36
C. 24
D. 12
E. 6

Simply use the fundamental counting principle discussed in video here:


and blog post here: https://anaprep.com/combinatorics-basic ... angements/

G's place is fixed: ___ ____ G ____ ____
H must be placed at either the 1st or the 5th spots i.e. there are 2 spots for H.
Rest of the 3 letters will be arranged in 3 spots in 3! ways

Total arrangements = 2 * 3! = 12

Answer (D)
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