The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT

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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

Answer: D
Source: Official guide
Source: — Problem Solving |

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BTGModeratorVI wrote:
Mon Apr 13, 2020 3:41 pm
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

Answer: D
Source: Official guide
Total no of ways arranging 5 letter with one letter redundant is \(\dfrac{5!}{2!}=60\)
No of ways two I's can be together \(4!=24\)

Number of ways at least one alpha is between two I's \(=60-24=36\)

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BTGModeratorVI wrote:
Mon Apr 13, 2020 3:41 pm
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

Answer: D
Source: Official guide
Another approach.

Take the task of arranging the 5 letters and break it into stages.

Stage 1: Arrange the 3 CONSONANTS (D, G and T) in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 consonants in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways

IMPORTANT: For each arrangement of 3 consonants, there are 4 places where the two I's can be placed.
For example, in the arrangement DTG, we can add spaces as follows _D_T_G_
So, if we place each I in one of the available spaces, we can ENSURE that the two I's are never together.

Stage 2: Select two available spaces and place an I in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)

Answer: D

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BTGModeratorVI wrote:
Mon Apr 13, 2020 3:41 pm
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

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.

We also have the following equation:

60 = (number of ways to arrange the letters with the I’s together) + (number of ways without the I’s together).

Let’s determine the number of ways to arrange the letters with the I’s together. Note that we treat the two I’s as a single item, so we are in essence finding the number of permutations of 4 items.

We have: [I-I] [D] [G] [T]

We see that with the I’s together, we have 4! = 24 ways to arrange the letters.

Thus, the number of ways to arrange the letters without the I’s together (i.e., with the I’s separated) is 60 - 24 = 36.

Answer: D

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