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D
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Difficulty:
45%
(medium)
Question Stats:
68% (01:42) correct
32%
(02:00)
wrong
based on 10303
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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
A) 12
B) 18
C) 24
D) 36
E) 48
15 Jun 2016, 02:55
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IMO 36
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24
no of ways at least one alpha is between two I's =60-24=36
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24
no of ways at least one alpha is between two I's =60-24=36
20 May 2017, 04:56
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skysailor
THEORY:
Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).
For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
For more check the links below:
Combinatorics Made Easy!
Theory on Combinations
DS questions on Combinations
PS questions on Combinations
Tough and tricky questions on Combinations
So, as explained above, the number of arrangements of 5 letters D, G, I, I , and T out of which there are two identical I's will be 5!/2! = 60.
