aarushisingla wrote:
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Hi,
Could you please explain why the below method is incorrect as I could not find by mistake
- If the 2 I's are separated by 1 letter
3! * 4C2 because there are 4 gaps between D, G and I = 36
- If the 2 I's are separated by 2 letters
3! * 3C2 because now there are 3 gaps. = 18
- If the 2 I's are separated by 3 letters
3! = 6
Total = 60.
Hi aarushisingla,
Based on your approach, it does not look like you are properly accounting for 'duplicate' entries.
For example:
D - (first I) - G - (second I) - T
and
D - (second I) - G - (first I) - T
are the SAME result, so you are not supposed to count it twice (just once).
I'm going to approach the first part of your calculation in a different way - and then you can try to complete the overall calculation on your own:
- If the 2 I's are separated by 1 letter
There are only three possible placements for the two Is:
I _ I _ _
_ I _ I _
_ _ I _ I
You can arrange the remaining 3 letters (the D, G and T) in each option in 3! = 6 ways. So the total number of ways to have the two Is separated by just 1 space is 3(3!) = 18.
GMAT assassins aren't born, they're made,
Rich
Ohh yes, I got it.
Thankyouu so much for help.
17 May 2020, 14:06
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