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

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07 Mar 2018, 11:31
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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 possibilities of forming 5 letter words with alphabets D I G I T is 5*4*3*2*1

Here, there are 5 possibilities for the first position, 4 for the second position, 3 for the third position,
2 for the second position, and 1 for the final position

However, since we have 2 I's, we divide the total possibilities by 2

Total possibilities with digits D I G I T are $$\frac{5*4*3*2*1}{2} = 60$$

Possibilities when the two I's are together
Consider the 2 I's as 1 unit. Let's call it X.
For the digits X D G T, Total possibilities are 4*3*2*1 = 24

Possibilities where I's are not together = Total possibilities - possibilities when they come together = 60-24 = 36

Therefore, the total possibilities in which the two occurrences of I aren't together is 36(Option 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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28 Mar 2018, 22:24
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

Leftmost arrangement of constraints:I_I_ _

Using formula, we have the number of permutations as (2!/2!) * 3!* 6=36

For explanation of the formula see link below and look for link Permutation with constraints.
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The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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01 May 2018, 05:04
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

hello again pushpitkc, generis, niks18

Here is my solution, its a bit different unlike others i replaced D,G,I,T, I with following letters $$A$$, $$B$$, $$C$$, $$D$$, $$E$$, so here, i simply modified question a bit yet arrived at the correct solution.

Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letters A and E are separated by at least one other letter?

So here is my reasoning

$$A$$, $$B$$, $$C$$, $$D$$, $$E$$

Step One Fix $$E$$, and count number of ways of $$A$$ which is 3! * 2 = 12 (we mutiply by 2, because A can be on either left or right side)

Step Two Fix $$A$$, and count number of ways of $$E$$ which is 3! * 2 = 12 (we mutiply by 2, because E can be on either left or right side)

Step Three Now we have to count number of ways of these three letters $$B$$, $$C$$, $$D$$ which is 3! * 2 = 12 ( mutiply by 2, because ORDER MATTERS/ PERMUTATION)

So total number of ways is $$12+12+12 = 36$$

i doubt that my third step is correct...though ... any ideas ? Quick tips? please

pushpitkc, generis niks18 did you see this post
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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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02 May 2018, 09:24
1
dave13 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

hello again pushpitkc, generis, niks18

Here is my solution, its a bit different unlike others i replaced D,G,I,T, I with following letters $$A$$, $$B$$, $$C$$, $$D$$, $$E$$, so here, i simply modified question a bit yet arrived at the correct solution.

Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letters A and E are separated by at least one other letter?

So here is my reasoning

$$A$$, $$B$$, $$C$$, $$D$$, $$E$$

Step One Fix $$E$$, and count number of ways of $$A$$ which is 3! * 2 = 12 (we mutiply by 2, because A can be on either left or right side)

Step Two Fix $$A$$, and count number of ways of $$E$$ which is 3! * 2 = 12 (we mutiply by 2, because E can be on either left or right side)

Step Three Now we have to count number of ways of these three letters $$B$$, $$C$$, $$D$$ which is 3! * 2 = 12 ( mutiply by 2, because ORDER MATTERS/ PERMUTATION)

So total number of ways is $$12+12+12 = 36$$

i doubt that my third step is correct...though ... any ideas ? Quick tips? please

pushpitkc, generis niks18 did you see this post

Hi dave13

I am finding it hard to understand your approach

However without complicating the matter, the question can be simply solved as

Total Number of Arrangements Possible - Number of arrangements where two I's are always together = Number of ways where two I's are never together.

You simply need to calculate the values of the above equation. Refer to pushpitkc 's solution above for clarity
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The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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09 May 2018, 10:52
Hi Bunuel, can I check why the no of ways when the Is are placed together is 4!? It can be located _D_G_T_. Why isn't it 4 ways? Thanks in advance!
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Posts: 47983
Re: The letters D, G, I, I , and T can be used to form 5-letter strings as  [#permalink]

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10 May 2018, 01:18
roastedchips wrote:
Hi Bunuel, can I check why the no of ways when the Is are placed together is 4!? It can be located _D_G_T_. Why isn't it 4 ways? Thanks in advance!

In addition to that D, G, and T, could be arranged in 3! ways, so total = 4*3! = 4!.

Or, consider two I's as one unit {II}. We'll have 4 units: {D}, {G}, {T}, and {II}. The number of arrangements is 4!.

Hope it's clear.
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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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05 Jul 2018, 18:55
Quote:
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.

In this stage, why did we have to select 2 spaces and not just place the 1st "I" (4 ways) and the 2nd "I" (3 ways)? For a total of 12 ways?
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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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09 Jul 2018, 04:56
Leo8 wrote:
Attachment:
FullSizeRender-3.jpg

Ans :D

Excellent explanation! thank you for your support
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as &nbs [#permalink] 09 Jul 2018, 04:56

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