Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Send along your receipt from another course or book to info@empowergmat.com and EMPOWERgmat will give you 50% off the first month of access OR $50 off the 3 Month Plan Only available to new students Ends: June 18th
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.
Re: In how many ways can the letters of the word JUPITER be arranged in a
[#permalink]
Show Tags
20 May 2011, 11:10
2
1
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 E-I-U-- with one letter spacing in between = 3(positions of EIU shifting) * 4! (letters to be arranged) = 72
4. Consider E-I--U- and E---I-U 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).
Re: In how many ways can the letters of the word JUPITER be arranged in a
[#permalink]
Show Tags
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!.
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 706
Re: In how many ways can the letters of the word JUPITER be arranged in a
[#permalink]
Show Tags
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!.
Re: In how many ways can the letters of the word JUPITER be arranged in a
[#permalink]
Show Tags
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
_________________
If you liked my post, please consider a Kudos for me. Thanks!
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
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
In how many ways can the letters of the word JUPITER be arranged in a
[#permalink]
Show Tags
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...:-(
Affiliations: National Institute of Technology, Durgapur
Joined: 22 Feb 2017
Posts: 30
Location: India
GMAT 1: 720 Q49 V38
GPA: 3.6
WE: Engineering (Manufacturing)
In how many ways can the letters of the word JUPITER be arranged in a
[#permalink]
Show Tags
23 Jan 2019, 00:07
1
1
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 device
If 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 '7-seats' 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 '7-seats' => We have to choose '3-seats' 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 '4-seats' for consonants (J, P,T, & R) => We can choose '4-seats' 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...
Re: In how many ways can the letters of the word JUPITER be arranged in a
[#permalink]
Show Tags
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 device
If 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 '7-seats' 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 '7-seats' => We have to choose '3-seats' 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 '4-seats' for consonants (J, P,T, & R) => We can choose '4-seats' 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.
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
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
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
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.
Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Re: In how many ways can the letters of the word JUPITER be arranged in a
[#permalink]
Show Tags
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 E-I-U-- with one letter spacing in between = 3(positions of EIU shifting) * 4! (letters to be arranged) = 72
4. Consider E-I--U- and E---I-U 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 E-I-U-- with one letter spacing in between = 3(positions of EIU shifting) * 4! (letters to be arranged) = 72
4. Consider E-I--U- and E---I-U 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).
close
69% (02:12) correct 
:-(
this was so helpful
