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Anjummm1
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The letters D, G, I, I , and T can be used to form 5-letter strings as 26 Nov 2024, 04:40
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Can someone please explain why is the number of ways in which Is are together is 24?
I thought it would be 4!*2! since the two I's can also explain positions and that would give us more possibilities

Please explain if I got it wrong
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Bunuel
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The letters D, G, I, I , and T can be used to form 5-letter strings as 26 Nov 2024, 04:51
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Anjummm1
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

Can someone please explain why is the number of ways in which Is are together is 24?
I thought it would be 4!*2! since the two I's can also explain positions and that would give us more possibilities

Please explain if I got it wrong

Consider the two I's as a single unit {II}. This creates 4 distinct units: {II}, {D}, {G}, and {T}. The number of ways to arrange these units is 4! = 24. The two I's within the unit are identical and can only be arranged in one way: {II}.

P.S. I’d recommend reviewing the earlier pages of the discussion for more insights.
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The letters D, G, I, I , and T can be used to form 5-letter strings as 11 Jan 2025, 04:45
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Hi Svaidyaraman,

what is the logic behind 3!*4 to figure out the number of permutations where I's are together? I'd appreciate your insight

Thank you.
SVaidyaraman
Bunuel
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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The letters D, G, I, I , and T can be used to form 5-letter strings as 11 Jan 2025, 04:57
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When the two I’s must stay together, we can treat them as a single unit or "block." This reduces the number of elements to arrange from 5 letters to 4 units => {[I,I], D, G, T}
You can either directly take 4! or 3!*4
The logic behind 3!*4 is =>
3!: The number of ways to arrange the other three letters (D, G, T) outside of the [I,I] block.
And, 4: The number of possible positions for the [I,I] block among the 4 positions in the arrangement.

Hope it helps!

BelisariusTirto
Hi Svaidyaraman,

what is the logic behind 3!*4 to figure out the number of permutations where I's are together? I'd appreciate your insight

Thank you.
SVaidyaraman
Bunuel
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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The letters D, G, I, I , and T can be used to form 5-letter strings as 11 Jan 2025, 05:19
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That clears it up for me, thank you! I appreciate you explaining why '4' is there too (even if it is probably obvious, given 3! and that there is 4 places I I can go in between them, but you left no room for confusion).

Krunaal
When the two I’s must stay together, we can treat them as a single unit or "block." This reduces the number of elements to arrange from 5 letters to 4 units => {[I,I], D, G, T}
You can either directly take 4! or 3!*4
The logic behind 3!*4 is =>
3!: The number of ways to arrange the other three letters (D, G, T) outside of the [I,I] block.
And, 4: The number of possible positions for the [I,I] block among the 4 positions in the arrangement.

Hope it helps!
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The letters D, G, I, I , and T can be used to form 5-letter strings as 28 Jan 2025, 12:43
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think two I's as 1 item since they will not break apart. then you will have 4 items (II,D,G,T) which makes 4!=24 different ways of permutation.
A77777
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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

can you please explain why the No of ways two I's can be together is 4!=24?
thank you!!
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The letters D, G, I, I , and T can be used to form 5-letter strings as 23 Feb 2025, 13:18
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Bunuel the problem does not say DISTINCT 5-letter strings, why are we dividing by 2!.

This problem should be 5! - 4!. as it should count for the duplicates.
Bunuel
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
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The letters D, G, I, I , and T can be used to form 5-letter strings as 23 Feb 2025, 13:33
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INprimesItrust
Bunuel the problem does not say DISTINCT 5-letter strings, why are we dividing by 2!.

This problem should be 5! - 4!. as it should count for the duplicates.
Bunuel
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

The answer should be and is 5!/2! - 4! = 36. The letters D, G, I, I, and T can be arranged in 5!/2! ways because there are two repeated I’s. The term “distinct” is implicit and doesn’t need to be stated explicitly—it’s the only correct interpretation. In other words, there is no difference between DGI2I1 and DGI2I1. This question has generated a three-page discussion, so you’d benefit from reading and studying it carefully.
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