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

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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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New post 14 Feb 2019, 08:42
# of ways in which I's are together: 4! (glue method). Usually with the glue method need to multiply by 2, but given I's are the same, don't need to. How many total ways can "digit" be arranged with no restriction? 5! = 120. Need to divide by 2! given the repetition. 60-24 = 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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New post 15 Feb 2019, 00:47
chetan2u
In the 2nd case ie when the 2 Is are together why are we not dividing the 4! by 2!, as the interchange of the 2 Is when they are together will lead to double the number of arrangements...
Please correct me...
Thanks in advance
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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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New post 10 Mar 2019, 20:45
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The correct answer is Choice D.

First, determine the total number of ways of rearranging the letters in DIGIT.

5! / 2! (accounts for the repetition of the 2 letter "I"s) = (5 x 4 x 3 x 2) / 2 = 120 / 2 = 60

Next, consider all the ways two letter "I"s can be adjacent within the word. They can either be in the 1/2 slot, the 2/3 slot, the 3/4 slot, or the 4/5 slot (4 locations), and for each of those 4 locations there are 3 x 2 x 1 = 6 other ways of rearranging the final 3 letters, so multiply 6 by 4 to get 24.

Subtract those 24 instances from the total to get 60 - 24 = 36

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New post 04 Jun 2019, 19:23
total number of ways that the letters can be arranged is (5!/2!) =60

it is easiest to find how many ways the I's can be together.

First we can treat both I's as a single entity ( I &I ).

This essentially this means we are arranging four entities which becomes 4!.

Thus the answer to the question is (5!/2!)- 4! = 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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New post 22 Jun 2019, 06:39
@
adiagr 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


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

60 - 24 = 36.

D is the answer.


I can never get this right! I always end up putting 4!*2! when i have to calculate ways of arranging DIGIT with both th I's combined, that because in my head i think both the I's can also be arranged in two ways. How do i get to correct this ! Bunuel Please help
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New post Updated on: 07 Aug 2019, 22:03
We have letters: D, G, I, I, T
We need to have at least one letter between both the "I"s.
We can find the answer by finding: (Total Number of ways the 5 letters can be arranged) - (Number of ways the letters are arranged such that both the "I"s stay together) -> (a)

Number of ways the 5 letters can be arranged is \(\frac{5!}{2!}\) = 60 ways

To calculate the number of ways the letters can be arranged such that both the "I"s stay together, we need to consider both "I"s as one element (can't be separated). Thus, we only have 4 elements to arrange now- "D", "G", "II", "T".
These 4 elements can be arranged in 4! ways = 24 ways [We do not have to worry about interchanging the two "I"s positions are they are not distinct elements].

Thus, From (a), the required number of arrangements = 60 - 24 = 36 ways.

Answer D

Originally posted by Sayon on 01 Aug 2019, 21:03.
Last edited by Sayon on 07 Aug 2019, 22:03, edited 1 time in total.
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New post 04 Aug 2019, 21:55
one of the easiest. We need to remember that 'Not together' in Permutation and Combinations can be achieved by
Total arrangements/combinations(Minus)together.
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New post 17 Aug 2019, 00:19
varundixitmro2512 wrote:
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


Why isn’t 4 factorial divided by 2 factorial?

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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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New post 17 Aug 2019, 01:17
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


Given: The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT.

Asked: 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?

Total 5-letters strings that can be formed by using {D,G,I,I,T} = 5!/2! = 60
5-letter strings that can be formed by using II together = 4! = 24
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

IMO D
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New post 22 Sep 2019, 22:58
ScottTargetTestPrep 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


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


Hi, Scott,

I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).

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New post 20 Nov 2019, 16:58
varundixitmro2512 wrote:
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


can you please explain why the No of ways two I's can be together is 4!=24?
thank you!!
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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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New post 20 Nov 2019, 21:08
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Hi A77777,

For the sake of calculating a permutation, the 'restriction' that the two "I"s must be 'next to one another' is essentially the same as saying that there is only one "I."

With 4 different letters, there are 4! = (4)(3)(2)(1) = 24 different arrangements

You can actually 'map' them all out if you choose; rather than list every option all at once, here are the 6 options that would start-off with the Is:

II D G T
II D T G
II G D T
II G T D
II T D G
II T G D

... and the 6 options that would start-off with the D:

D G II T
D G T II
D II G T
D II T G
D T G II
D T II G

There are two other 'sets of 6'; one that starts with a G and another that starts with a T.

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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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New post 20 Nov 2019, 22:27
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Hi, EmpowerGMATRich,

I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).

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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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New post 20 Nov 2019, 22:36
Hi Raxit85,

You bring up a good question. Although the prompt does not explicitly state it, we are meant to assume that the two "I"s are identical. In simple terms, if you had just two letters (re: the two "I"s), then you could form just ONE distinct 'word': II (not two words - again, because the two "I"s are identical).

IF... the prompt did not want us to think in those terms, then we wouldn't have duplicate letters at all - it would just be 5 different letters and the question would ask something to the effect of "how many different ways are there to arrange the letters A, B, C, D and E so that there is at least one letter between the A and the B?"

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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as   [#permalink] 20 Nov 2019, 22:36

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