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The letters D, G, I, I , and T can be used to form 5letter strings as
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07 Mar 2018, 10:31
Bunuel wrote: The letters D, G, I, I , and T can be used to form 5letter strings as DIGIT or DGIIT. Using these letters, how many 5letter 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 = 6024 = 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 5letter strings as
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09 Mar 2018, 00:26
pushpitkc wrote: Bunuel wrote: The letters D, G, I, I , and T can be used to form 5letter strings as DIGIT or DGIIT. Using these letters, how many 5letter 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 = 6024 = 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*34*3*2*1 =4*3(52) =12(3)=36



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Re: The letters D, G, I, I , and T can be used to form 5letter strings as
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28 Mar 2018, 21:24
Bunuel wrote: The letters D, G, I, I , and T can be used to form 5letter strings as DIGIT or DGIIT. Using these letters, how many 5letter 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 5letter strings as
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01 May 2018, 04:04
Bunuel wrote: The letters D, G, I, I , and T can be used to form 5letter strings as DIGIT or DGIIT. Using these letters, how many 5letter 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 5letter 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 5letter strings as
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02 May 2018, 08:24
dave13 wrote: Bunuel wrote: The letters D, G, I, I , and T can be used to form 5letter strings as DIGIT or DGIIT. Using these letters, how many 5letter 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 5letter 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 5letter strings as
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09 May 2018, 09:52
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 5letter strings as
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10 May 2018, 00:18
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 5letter strings as
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05 Jul 2018, 17:55
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 5letter strings as
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09 Jul 2018, 03:56
Leo8 wrote: Attachment: FullSizeRender3.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 5letter strings as
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09 Sep 2018, 22:33
ScottTargetTestPrep wrote: Bunuel wrote: The letters D, G, I, I , and T can be used to form 5letter strings as DIGIT or DGIIT. Using these letters, how many 5letter 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: [II] [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 When I's are together , isn't the equation be 4!/2!? identical I's should also be arranged in 2! ways..Please explain?



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Re: The letters D, G, I, I , and T can be used to form 5letter strings as
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10 Nov 2018, 12:19
I do understand why we divided 5! by 2! to get total arrangements before constraint but I dont understand why we also didnt divide 4! by 2! while calculating the constraints (since both IIs will be the same in DIIGT for example)? Could someone help me with that? Thanks



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Re: The letters D, G, I, I , and T can be used to form 5letter strings as
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06 Dec 2018, 13:25
Bunuel wrote: The letters D, G, I, I , and T can be used to form 5letter strings as DIGIT or DGIIT. Using these letters, how many 5letter 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 When answering counting question, one should always consider whether a useful approach would be listing and counting the possible outcomes. How do we know when this approach may be the best approach? If you check the answer choices (ALWAYS check the answer choices BEFORE performing any calculations) and find that the answer choices are relatively small (as they are here), then listing and counting might be a great approach. Alternatively, if you have no idea how to solve the question using traditional counting techniques, you might again consider listing and counting. Also, even if the answer choices are relatively large, you might start listing some outcomes, and while doing so, you might see a pattern emerge that you can use to determine the correct answer. TLDR: Don't dismiss listing and counting as a possible approach to answering counting questions. POSSIBLE OUTCOMESWords in the form _ I _ I _ DIGIT DITIG GITID GIDIT TIGID TIDIG Words in the form _ I _ _ I DIGTI DITGI GIDTI GITDI TIDGI TIGDI Words in the form _ _ I _ I DGITI DTIGI GDITI GTIDI TDIGI TGIDI Words in the form I _ I_ _ITIGD ITIDG IDITG IDIGT IGITD IGIDT Words in the form I _ _ I_ IDGIT IGDIT ITGID IGTID IDTIG ITDIG Words in the form I _ _ _ I IDGTI IDTGI IGDTI IGTDI ITDGI ITGDI Done! Answer = 36 = D Did that take under 2 minutes? Probably. Cheers, Brent
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The letters D, G, I, I , and T can be used to form 5letter strings as
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19 Dec 2018, 05:37
Total no of ways arranging 5 letters when two letters are same is 5!/2! = 60 ways. No of ways two I's are always together is 4!=24 ways Hence required number of ways = 6024=36 ways (Option D)
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Re: The letters D, G, I, I , and T can be used to form 5letter strings as
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22 Dec 2018, 03:37
adiagr wrote: Bunuel wrote: The letters D, G, I, I , and T can be used to form 5letter strings as DIGIT or DGIIT. Using these letters, how many 5letter 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. Hi Bunuel. To count when I's always come together = Shouldn't it be 4! * 2! ? Because those two I's can be interchanged?




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