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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?

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15 Jun 2016, 03:00

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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

Let us calculate the total ways. Those would be (5!/2!) = 60.

Now since the question says "at least" let us find the number of arrangements when both I's are together. (Tie them up). so we have 4! ways to arrange such that I's always come together. 4! = 24

Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

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21 Jun 2016, 08:59

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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

Total ways to arrange these letters=5!/2!

Ways to arrange the letters if both I are together= 4!

Ways to arrange with both Is not together= 5!/2!-4!= 36

D is the answer
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

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23 Jun 2016, 01:23

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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

Required: atleast one letter between two Is

Total cases = 5!/2! = 60 Total cases in which Is will be together = 4! = 24 {Treat both Is as one}

Hence total cases in which there is atleast one letter between the Is = 60 - 24 = 36

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28 Jul 2016, 04:26

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The requirement is 5 letter strings in which two I's are not together, they are separated by atleast one letter.

Consider this as below,

(Total number of 5 letter strings that can be formed) - (Total number of 5 letter strings in which two I's are together) =(5!/2!) - (4!) = 5*4*3 - 4*3*2 = 60 - 24 = 36

Finding total number of 5 letter strings - DIGIT = 5!/2! (2! because of the repetition of I's) Finding total number of 5 letter strings with I's together - II_ _ _ = 4!
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

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10 Jan 2017, 09:53

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

@bunnel .. Just for info wanted to know how should have the question solved if it was said "how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by one other letter " instead of the one given !! Please help !!

Are you asking what the answer would be if we replaced the "at least" in the question stem with "exactly"?

If so, then, for your hypothetical question, there would be only a few ways to place the I's. So a 'direct approach' would be an efficient way to solve the question. We'd have:

I _ I _ _ _ I _ I _ _ _ I _ I

For each of these options we have three open spaces to arrange the other three letters. So for each of the options we'd have:

(3 choices for the first open slot) * (2 choices for the second open slot) * (1 choices for the last open slot) = 3! = 6 ways to complete the option.

Overall we'd multiply the 3 options by the 6 ways to complete each of the options. So, in your hypothetical version of the question, your answer would be 3*6 = 18.

The "direct approach" video in my previous post might help with the logic here.

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

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)

Could someone help me understand how the total number of cases is 5!/2!? I don't understand the reasoning behind that.

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 \(\frac{9!}{4!3!2!}\).

So, ss 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. _________________

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

1. It is an ordering, therefore a permutation problem 2. There is a constraint 3. The constraint is an "Atleast One" constraint. so the opposite of it "none" is easier to find 4. Total number of permutations is 5!/2!=60 ways since I repeats twice 5. Number of permutations where I's are together is 3!*4 = 24 6. Permutations with constraint is (4)-(5)=36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

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24 Jul 2017, 10:37

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

Just want to share my approach:

1. In this case, the 3 letter D,G,T can be calculated easily : 3! = 6. 2. The problem is only the remanining I : which must place with separation. 3. The best way to calculate the "I" case is to draw it :

I _ I _ _ (1) I _ _ I _ (2) I _ _ _ I (3) _ I _ I _ (4) _ I _ _ I (5) _ _ I _ I (6)

As we can see, there are only 6 possibilies. 4. Calculate all = 6 X 6 = 36.

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

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 Is, 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 Is together) + (number of ways without the Is together).

Let’s determine the number of ways to arrange the letters with the Is together.

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

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

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

Answer: D
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

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21 Aug 2017, 00:57

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

Letters are DGIIT - 5 letters... Total no. of 5 letter string that can be formed = 5!/2! = 5*4*3 = 60

Total no. of 5 letter string that in which 2 I are together that can be formed = 4! = 4*3*2 = 24

So, 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter = 60 - 24 = 36

Since the two "I"s cannot be side-by-side, there are a limited number of ways to arrange the 5 letters. As such, with a little permutation math and some 'brute force', we can map out the possibilities:

_ _ _ _ _

If the first letter is an I, then the second letter CANNOT be an I (it would have to be one of the other 3 non-I letters)...

i 3

From here, any of the remaining letters can be in the 3rd spot. After placing one, either of the remaining two letters can be in the 4th spot and the last letter would be in the 5th spot...

i 3 3 2 1

This would give us (3)(3)(2)(1) = 18 possible arrangements with an I in the 1st spot.

If a non-I is in the 1st spot and an I is in the 2nd spot, then we have...

3 i _ _ _

A non-I would have to be in the 3rd spot, then either remaining letter could be 4th...

3 i 2 2 1

This would give us (3)(2)(2)(1) = 12 possible arrangements

Next, we could have two non-Is to start off, then Is in the 3rd and 5th spots...

3 2 i 1 i

This would give us (3)(2)(1) = 6 possible arrangements

There are no other options to account for, so we have 18+12+6 total arrangements.