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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
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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09 Mar 2018, 01:26
pushpitkc 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 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)I solved this by this method No case is possible with 2 I together so if we subtract total no. of cases with the cases of 2 I together than that will be our answer total cases are 5!/2! (total letters/repeated letters) together case is 4!*2!/2!(two i together counted as 1 then total letter is 4* both 2 can exchange their places/2 repeated no.) =4! 5*4*3-4*3*2*1 =4*3(5-2) =12(3)=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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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, niks18Here 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
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, niks18Here 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 dave13I 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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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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