Another approach:
Take the task of arranging the 7 digits and break it into stages.

We’ll begin with the most restrictive stages.

Stage 1: Place the letter G
Since the letter G must be placed in the 3rd position, we can complete stage 1 in 1 way

Stage 2: Place the letter H
So far we have: _ _ G _ _
Since H cannot be next to G, we can only place H in the 1st position or the 5th position
So we can complete this stage in 2 ways.

Stage 3: Place the letter I
We can place I wherever we wish.
There are 3 spaces remaining, so we can complete this stage in 3 ways.

Stage 4: Place the letter J
We can place I wherever we wish.
There are 2 spaces remaining, so we can complete this stage in 2 ways.

Stage 5: Place the letter K
There is 1 space remaining, so we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus arrange all 5 letters) in (1)(2)(3)(2)(1) ways (= 12 ways)

Answer: D

Cheers,
Brent
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Explanation

When the 5 letters are rearranged, G is to be listed in the 3rd position and H cannot be next to G, so there are only two possible positions for H: 1st and 5th.

Case 1: In the rearranged list, G is in the 3rd position, H is in the 1st position, and each of the remaining 3 letters can be in any of the remaining 3 positions. The number of ways these remaining 3 letters can be arranged is 3!, or 6. Thus the total number of rearrangements in case 1 is 6.

Case 2: In the rearranged list, G is in the 3rd position, H is in the 5th position, and each of the remaining 3 letters can be in any of the remaining 3 positions. Thus, as in case 1, the total number of rearrangements in case 2 is 6.

Thus there are 6 + 6, or 12, possible rearrangements, and the correct answer is Choice D.
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@sandy why can't we do like this
1] total no of arrangements 5!=120
2] then place H in 2nd and 4th position, so no of ways are 12
3] then subtract 120-12 = 108

why is this approach wrong?

G is not in the third place in your assumption. If G is in third place

1. total number of arrangements 4!=24 (Not 5! as G is fixed to 3rd place)
2] then place H in 2nd and 4th position, so no of ways are 12
3] then subtract 24-12 = 12

Now its OK
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I liked ur explanation.
My question:

if we have to put G in the 3rd then why couldn't we solve it like this : 4 2 1 1 1

the first blank I can use H I J K, the second I can use any of ( J I K ), then G , then any of the rest except H, then the remaining.
what is wrong with that answer?

total 12 ways(6 plus 6) with first way being placing the at first and in another way one can place H at the last.

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 there?

A. 60
B. 36
C. 24
D. 12
E. 6
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There are total 5 slots for the 5 letters.
- - - - -
G has to go in the 3rd slot
- - 1 - - because there is only 1 option i.e. 'G' for the 3rd slot
H cannot be on either side of G that leaves us with 3 letters I J K
- 3 1 2 - Among those 3 letters we chose 1 and that leaves us with 2 other letters.
After spot for three of the letters are finalized we have two letters left hence
2 3 1 2 1 = 12. Remember this is a permutation problem so order matters
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[

I'd typically solve this question the same way amorphous did.
However, since the answer choices are relatively small, it's not a crazy idea to consider LISTING AND COUNTING all possible outcomes (especially if you're not sure how to solve the question.

If we're take a systematic approach, it won't take long.
We get:

HIGJK
HIGKJ
HJGIK
HJGKI
HKGIJ
HKGJI
IJGKH
IKGJH
JIGKH
JKGIH
KIGJH
KJGIH

TOTAL = 12

Answer: D

Cheers,
Brent
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Merged similar topics.

Brent also explained it two times.

Use the search button before to post a question https://greprepclub.com/forum/search.php

We do HAVE ALL the official questions released by ETS in the entire GRE history.

No need to post once again

Required Arrangement: _ _ G _ _

Total ways of arranging these 5 alphabets with G at 3rd position = 4 x 3 x G x 2 x 1 = 24
Total ways in which G and H are always together = _ H G _ _ + _ _ G H _ = (3 x H x G x 2 x 1) + (3 x 2 x G x H x 1) = 6 + 6 = 12

Total ways in which G is at the 3rd place and H is not next to it = 24 -12 = 12

Hence, option D
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