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In how many ways can the letters of the word JUPITER be arranged in a
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20 May 2011, 06:55
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In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order? (A) 736 (B) 768 (C) 792 (D) 840 (E) 876  IIM CAT Level Question 
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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30 Apr 2017, 11:27
hussi9 wrote: In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
(A) 736 (B) 768 (C) 792 (D) 840 (E) 876
 IIM CAT Level Question  Though I got it wrong, Here is the best Method to solve this question. JUPITER > 7 Alphabets > 3 Vowels and 4 Consonants. # Arrangements of 3 Vowels = 3! = 6. # of Valid arrangements out of these = 1 For all the other arrangements, vowels will not be arranged alphabetically.Number of Legitimate Arrangements = Total Arrangements  Illegitimate arrangements. \(= 7!  \frac{5}{6}*7!\) \(= 7!*(1  \frac{5}{6})\) \(=7! * \frac{1}{6}\) = 840
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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20 May 2011, 08:12
three letters can be arranged in 3! ways. only one combination EIU is required.
7 letters can be arranged in 7! ways.
thus 7!/ 3! * 1 = 840.
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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20 May 2011, 08:51
@amit2k9, could you please explain this in more detail ?
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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20 May 2011, 11:10
1. Consider EIU as one. then total possible combination of words = 5* 4! = 120
2. Consider EI and IU together total possible combinations = [5 (positions for EI out of 7) * 5! (rest of the letters)] + [ 5 (positions for IU)* 5! (rest of the letters)]
Now in half of the combinations U will be ahead of EI and E will be ahead of IU.
So 1/2[5 (positions for EI out of 7) * 5! (rest of the letters)] +1/2 [ 5 (positions for IU)* 5! (rest of the letters)]
total will give = 300 + 300 = 600.
3. Consider EIU with one letter spacing in between = 3(positions of EIU shifting) * 4! (letters to be arranged) = 72
4. Consider EIU and EIU with two letter spacing between IU and EI = 2 * 4! = 48
thus totaling 120 + 600 + 72 + 48 = 840.
Alternately , Consider
all possible letter arrangements = 7!
only one combination out of 3! combinations possible for EIU. meaning 3 are taken in a group,but they are not alike. Means total combinations remain 7! and not 5! ( EIU together + 4 letters).
So, essentially it is 7P3 = Hence 7!/ 3! = 840



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Re: In how many ways can the letters of the word JUPITER be arranged in a
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20 May 2011, 13:51
Amt2k9,
Conceptwise, I did not get your 7!/3!. If the word was, Let's say JUPUTUR, number of ways the letter can be arranged is 7!/3! ways (total ! / identical !). You used the same concept where the letters are not identical. You mentioned that those are not alike, but still not clear on why we still can do 7!/3!.



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Re: In how many ways can the letters of the word JUPITER be arranged in a
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20 May 2011, 19:20
Darn ! I am not Amt2k9. I think it is pretty obvious that in 1/6th of the ways out of 7! the vowels will be ordered. This reasoning is parallel to the situation when we have 3 people lets say A B C sitting in a row and we want to know the ways in which they will be alphabetical order. There is just one way ie. 3!/6=1 or 3!/3! The division by 3! is NOT owned by the concept of weeding out the duplicates. We will apply it when we need it bellcurve wrote: Amt2k9,
Conceptwise, I did not get your 7!/3!. If the word was, Let's say JUPUTUR, number of ways the letter can be arranged is 7!/3! ways (total ! / identical !). You used the same concept where the letters are not identical. You mentioned that those are not alike, but still not clear on why we still can do 7!/3!.



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Re: In how many ways can the letters of the word JUPITER be arranged in a
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20 May 2011, 23:27
bellcurve wrote: Amt2k9,
Conceptwise, I did not get your 7!/3!. If the word was, Let's say JUPUTUR, number of ways the letter can be arranged is 7!/3! ways (total ! / identical !). You used the same concept where the letters are not identical. You mentioned that those are not alike, but still not clear on why we still can do 7!/3!. Since the order of vowels will always remain the same despite these occupying different positions > if we assume each vowel as X then our question is same as asking "arrange JPTRXXX" => in all 7!/3! ways => Choice (4) is the right answer
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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05 Aug 2016, 23:44
hussi9 wrote: In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
(A) 736 (B) 768 (C) 792 (D) 840 (E) 876
Commenting as I struggled getting my head around the concept. Kudos given above. There are \(7_P7 = 7!\) ways to arrange the 7 letters. For each of these combinations, there is a single valid arrangement of the vowels: \(\frac{1}{3_P3}\) \(7! \times \frac{1}{3!} = \frac{7!}{3!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 840 = D\)
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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05 Jul 2017, 05:43
hussi9 wrote: In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
(A) 736 (B) 768 (C) 792 (D) 840 (E) 876
 IIM CAT Level Question  Without limitation the words of JUPITER can be arranged in 7! ways.. Now there are 3 vowels I, E, U So these vowels can be arranged among themselves in 3! ways. We need to find the arrangement in vowels are in alphabetical order i.e. EIU which is only one out of 3! ways. Hence, total number of required arrangement = 7!/3! = 7*6*5*4 = 42 * 20 = 840 Answer D
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In how many ways can the letters of the word JUPITER be arranged in a
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09 Dec 2018, 01:58
I have really tried to understand the concepts applied here from all the explanations but I still don't.Can someone be kind enough to help with a link that explains these concepts clearly... :( Thanks. Posted from my mobile device



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In how many ways can the letters of the word JUPITER be arranged in a
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23 Jan 2019, 00:07
Kem12 wrote: I have really tried to understand the concepts applied here from all the explanations but I still don't.Can someone be kind enough to help with a link that explains these concepts clearly... :( Thanks. Posted from my mobile deviceIf you are struggling with the above methods like me, I'm going to explain to you my way of solving ; However I don't know whether it would be sufficient to clear your doubts or not ?! There are a TOTAL '7seats' for the alphabets of the letter 'JUPITER' out of which '3' are vowels & '4' are consonants ; The question asks us to find the total no. of ways in which vowels will be in a particular order (,i.e, alphabetical order & if it was a no. code maybe the question would have told ascending/descending order) ; Coming back to the question, we know there are '7seats' => We have to choose '3seats' for our vowels ( E, I, & U ) => we can do that in 7C3 ways ; Or, simply [7!/(3!x4!)] ways [This is what I meant => C C V V V C C, C V C V C V C, C C C V V V C, etc.] Now, the order of vowels is 'E I U' => Total no. of arrangements = (7C3 x 1) [And NOT (7C3 x 3!) Since, doing so we would include the arrangements 'UIE, UEI, IUE, IEU, EUI' along with 'EIU' which we don't want as per the question] Now, we are left with '4seats' for consonants (J, P,T, & R) => We can choose '4seats' for '4' consonants in 4C4 (or 1) way ; [But here we are allowed for arrangements, i.e, JEUITPR, JEUITRP, EUIJPRT, EUITPR, etc] ; Hence, we can do it in 4C4x4! ways ; THEREFORE, TOTAL NO. of WAYS IN WHICH VOWELS STAYS IN ALPHABETICAL ORDER IN 'JUPITER' = 7C3x1x4C4x4! = 840 ways = option D. Hope it helps...



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Re: In how many ways can the letters of the word JUPITER be arranged in a
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23 Jan 2019, 04:13
Lithium wrote: Kem12 wrote: I have really tried to understand the concepts applied here from all the explanations but I still don't.Can someone be kind enough to help with a link that explains these concepts clearly... :( Thanks. Posted from my mobile deviceIf you are struggling with the above methods like me, I'm going to explain to you my way of solving ; However I don't know whether it would be sufficient to clear your doubts or not ?! There are a TOTAL '7seats' for the alphabets of the letter 'JUPITER' out of which '3' are vowels & '4' are consonants ; The question asks us to find the total no. of ways in which vowels will be in a particular order (,i.e, alphabetical order & if it was a no. code maybe the question would have told ascending/descending order) ; Coming back to the question, we know there are '7seats' => We have to choose '3seats' for our vowels ( E, I, & U ) => we can do that in 7C3 ways ; Or, simply [7!/(3!x4!)] ways [This is what I meant => C C V V V C C, C V C V C V C, C C C V V V C, etc.] Now, the order of vowels is 'E I U' => Total no. of arrangements = (7C3 x 1) [And NOT (7C3 x 3!) Since, doing so we would include the arrangements 'UIE, UEI, IUE, IEU, EUI' along with 'EIU' which we don't want as per the question] Now, we are left with '4seats' for consonants (J, P,T, & R) => We can choose '4seats' for '4' consonants in 4C4 (or 1) way ; [But here we are allowed for arrangements, i.e, JEUITPR, JEUITRP, EUIJPRT, EUITPR, etc] ; Hence, we can do it in 4C4x4! ways ; THEREFORE, TOTAL NO. of WAYS IN WHICH VOWELS STAYS IN ALPHABETICAL ORDER IN 'JUPITER' = 7C3x1x4C4x4! = 840 ways = option D. Hope it helps... Thank you so much this was so helpful



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Re: In how many ways can the letters of the word JUPITER be arranged in a
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23 Jan 2019, 04:36
hussi9 wrote: In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
(A) 736 (B) 768 (C) 792 (D) 840 (E) 876
 IIM CAT Level Question  Alternatively, think of it this way: The vowels must be arranged like this: E I U Now there are 4 spots (shown by x) for J: x E x I x U x Say we choose E J I U Next we place P. There are 5 spots for P: x E x J x I x U x and so on for T and R too. So in all we get 4*5*6*7 = 840 different ways (you don't need to calculate this. Since there is a 2 and 5, it will end in 0) Answer (D)
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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23 Jan 2019, 04:50
hussi9 wrote: In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
(A) 736 (B) 768 (C) 792 (D) 840 (E) 876
 IIM CAT Level Question  JUPITER total 7 ! is the ways it can be arranged but question has asked vowels appear in alphabetical order so EIUJPTR would be the sequence we can put EIU ; 3! 7!/3! = 840 IMO D
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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24 Jan 2019, 20:15
VeritasKarishma wrote: hussi9 wrote: In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
(A) 736 (B) 768 (C) 792 (D) 840 (E) 876
 IIM CAT Level Question  Alternatively, think of it this way: The vowels must be arranged like this: E I U Now there are 4 spots (shown by x) for J: x E x I x U x Say we choose E J I U Next we place P. There are 5 spots for P: x E x J x I x U x and so on for T and R too. So in all we get 4*5*6*7 = 840 different ways (you don't need to calculate this. Since there is a 2 and 5, it will end in 0) Answer (D) Hi VeritasKarishma, Kindly, help me understand why P has 5 spots and so on. J has 4 spots and then should't be 3 spots for P and so on... Please help!



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In how many ways can the letters of the word JUPITER be arranged in a
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hussi9 wrote: In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
(A) 736 (B) 768 (C) 792 (D) 840 (E) 876
 IIM CAT Level Question  I solved this question,in the inline way i fixed the position of E E _ _ _ _ _ _, Now here if I U takes the 1st 2 spots Then the remaining spots will be filled in 4*3*2*1 = 24 ways and I U can move 4 more times making the total ways as 24*5 = 120 ways In the next case I will fix E I this time and move the rest of the alphabets i.e. E I _ U _ _ _ and so on, this will give you 96 ways. Next Case, I fixed E I U, this will give you 24 ways. Total = 120 + 96 + 24= 240, now EIU can also be UIE, this makes 240*2 ways = 840 ways OA D
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Re: In how many ways can the letters of the word JUPITER be arranged in a
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24 Jan 2019, 20:44
amit2k9 wrote: 1. Consider EIU as one. then total possible combination of words = 5* 4! = 120
2. Consider EI and IU together total possible combinations = [5 (positions for EI out of 7) * 5! (rest of the letters)] + [ 5 (positions for IU)* 5! (rest of the letters)]
Now in half of the combinations U will be ahead of EI and E will be ahead of IU.
So 1/2[5 (positions for EI out of 7) * 5! (rest of the letters)] +1/2 [ 5 (positions for IU)* 5! (rest of the letters)]
total will give = 300 + 300 = 600.
3. Consider EIU with one letter spacing in between = 3(positions of EIU shifting) * 4! (letters to be arranged) = 72
4. Consider EIU and EIU with two letter spacing between IU and EI = 2 * 4! = 48
thus totaling 120 + 600 + 72 + 48 = 840.
Alternately , Consider
all possible letter arrangements = 7!
only one combination out of 3! combinations possible for EIU. meaning 3 are taken in a group,but they are not alike. Means total combinations remain 7! and not 5! ( EIU together + 4 letters).
So, essentially it is 7P3 = Hence 7!/ 3! = 840 amit2k9 wrote: 1. Consider EIU as one. then total possible combination of words = 5* 4! = 120
2. Consider EI and IU together total possible combinations = [5 (positions for EI out of 7) * 5! (rest of the letters)] + [ 5 (positions for IU)* 5! (rest of the letters)]
Now in half of the combinations U will be ahead of EI and E will be ahead of IU.
So 1/2[5 (positions for EI out of 7) * 5! (rest of the letters)] +1/2 [ 5 (positions for IU)* 5! (rest of the letters)]
total will give = 300 + 300 = 600.
3. Consider EIU with one letter spacing in between = 3(positions of EIU shifting) * 4! (letters to be arranged) = 72
4. Consider EIU and EIU with two letter spacing between IU and EI = 2 * 4! = 48
thus totaling 120 + 600 + 72 + 48 = 840.
Alternately , Consider
all possible letter arrangements = 7!
only one combination out of 3! combinations possible for EIU. meaning 3 are taken in a group,but they are not alike. Means total combinations remain 7! and not 5! ( EIU together + 4 letters).
So, essentially it is 7P3 = Hence 7!/ 3! = 840 Sent from my Moto G (5) Plus using GMAT Club Forum mobile app




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