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Start the planA) 12
B) 18
C) 24
D) 36
E) 48
Answer: D
Source: Official guide
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Start the planTotal no of ways arranging 5 letter with one letter redundant is \(\dfrac{5!}{2!}=60\)BTGModeratorVI wrote: ↑Mon Apr 13, 2020 3:41 pmThe 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
Answer: D
Source: Official guide
Another approach.BTGModeratorVI wrote: ↑Mon Apr 13, 2020 3:41 pmThe 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
Answer: D
Source: Official guide
This is a permutation problem because the order of the letters matters. Let’s first determine in how many ways we can arrange the letters. Since there are 2 repeating I’s, we use the indistinguishable permutations formula, and we see that we can arrange the letters in 5! / 2! = 120 / 2 = 60 ways.BTGModeratorVI wrote: ↑Mon Apr 13, 2020 3:41 pmThe 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
Answer: D
Source: Official guide
