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:
Quant Quizzes are back with a Bang and with lots of Prizes. The first Quiz will be on 8th Dec, 6PM PST (7:30AM IST). The Quiz will be Live for 12 hrs. Solution can be posted anytime between 6PM-6AM PST. Please click the link for all of the details.
Join IIMU Director to gain an understanding of DEM program, its curriculum & about the career prospects through a Q&A chat session. Dec 11th at 8 PM IST and 6:30 PST
Enter The Economist GMAT Tutor’s Brightest Minds competition – it’s completely free! All you have to do is take our online GMAT simulation test and put your mind to the test. Are you ready? This competition closes on December 13th.
Attend a Veritas Prep GMAT Class for Free. With free trial classes you can work with a 99th percentile expert free of charge. Learn valuable strategies and find your new favorite instructor; click for a list of upcoming dates and teachers.
Does GMAT RC seem like an uphill battle? e-GMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days.
The letters C, I, R, C, L, and E can be used to form 6-letter strings
[#permalink]
Show Tags
21 Sep 2019, 03:16
13
00:00
A
B
C
D
E
Difficulty:
45% (medium)
Question Stats:
70% (02:05) correct 30% (02:00) wrong based on 207 sessions
HideShow timer Statistics
The letters C, I, R, C, L, and E can be used to form 6-letter strings such as CIRCLE or CCIRLE. Using these letters, how many different 6-letter strings can be formed in which the two occurrences of the letter C are separated by at least one other letter?
Re: The letters C, I, R, C, L, and E can be used to form 6-letter strings
[#permalink]
Show Tags
21 Sep 2019, 03:25
6
5
gmatt1476 wrote:
The letters C, I, R, C, L, and E can be used to form 6-letter strings such as CIRCLE or CCIRLE. Using these letters, how many different 6-letter strings can be formed in which the two occurrences of the letter C are separated by at least one other letter?
A. 96 B. 120 C. 144 D. 180 E. 240
PS54110.01
Good arrangements = total arrangements - bad arrangements.
Total arrangements: Number of ways to arrange 6 elements = 6!. But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED. The reason: When the identical elements swap positions, the arrangement doesn't change. Here, we must divide by 2! to account for the two identical C's: 6!/2! = 360.
Bad arrangements: In a bad arrangement, the two C's are in adjacent slots. Let [CC] represent the 2 adjacent C's. Number of ways to arrange the 5 elements [CC], I, R, L and E = 5! = 120.
Good arrangements: Total arrangements - bad arrangements = 360-120 = 240.
Re: The letters C, I, R, C, L, and E can be used to form 6-letter strings
[#permalink]
Show Tags
21 Sep 2019, 05:18
2
Total possible arrangements for CIRCLE ; 6!/2! ; 360 and arrangements for CCIRLE ; CC; X ; XIRLE; 5! ; 120 so different 6-letter strings can be formed in which the two occurrences of the letter C are separated by at least one other letter = 360-120 ; 240 IMO E
gmatt1476 wrote:
The letters C, I, R, C, L, and E can be used to form 6-letter strings such as CIRCLE or CCIRLE. Using these letters, how many different 6-letter strings can be formed in which the two occurrences of the letter C are separated by at least one other letter?
The letters C, I, R, C, L, and E can be used to form 6-letter strings
[#permalink]
Show Tags
26 Oct 2019, 09:54
4
Top Contributor
gmatt1476 wrote:
The letters C, I, R, C, L, and E can be used to form 6-letter strings such as CIRCLE or CCIRLE. Using these letters, how many different 6-letter strings can be formed in which the two occurrences of the letter C are separated by at least one other letter?
A. 96 B. 120 C. 144 D. 180 E. 240
We can use the rule that says: TOTAL number of outcomes if we IGNORE the rule = (number of outcomes that FOLLOW the rule) + (number of outcomes that BREAK the rule) In other words: Number of ways to arrange the 6 letter if we IGNORE the rule = (number of words that DON'T have adjacent C's) + (number of words that DO have adjacent C's)
Rearrange to get: number of words that DON'T have adjacent C's = (Number of ways to arrange the 6 letter if we IGNORE the rule) - (number of words that DO have adjacent C's)
Number of ways to arrange the 6 letter if we IGNORE the rule If we IGNORE the rule, then we are arranging the letters in CIRCLE Since we have DUPLICATE letters, we can apply the MISSISSIPPI rule (see video below)
In the word CIRCLE: There are 6 letters in total There are 2 identical C's So, the total number of possible arrangements = 6!/(2!) = 360
number of words that DO have adjacent C's Take the two C's and "glue" them together to get the SUPER LETTER "CC" This ensures that the C's are together We now must arrange CC, I, R, L, E We can arrange n different objects in n! ways So, we can arrange CC, I, R, L, and E in 5! ways (= 120 ways) So, number of words that DO have adjacent C's = 120
So, number of words that DON'T have adjacent C's = 360 - 120 = 240
Re: The letters C, I, R, C, L, and E can be used to form 6-letter strings
[#permalink]
Show Tags
26 Oct 2019, 15:33
gmatt1476 wrote:
The letters C, I, R, C, L, and E can be used to form 6-letter strings such as CIRCLE or CCIRLE. Using these letters, how many different 6-letter strings can be formed in which the two occurrences of the letter C are separated by at least one other letter?
A. 96 B. 120 C. 144 D. 180 E. 240
PS54110.01
Letters in word CIRCLE can be arranged in !6/!2 ways, and for the said condition we have to subtract those cases(bundle those two C's together and count them as 1 unit) where two C's are coming together. !6/!2-!5 =240 E:)
Re: The letters C, I, R, C, L, and E can be used to form 6-letter strings
[#permalink]
Show Tags
31 Oct 2019, 08:13
Total number of combinations possible= 6!/2=360 Number of combinations with 2 Cs together = 5!=120 Number of combinations with 2 Cs separated = 360-120= 240
Re: The letters C, I, R, C, L, and E can be used to form 6-letter strings
[#permalink]
Show Tags
23 Nov 2019, 10:25
gmatt1476 wrote:
The letters C, I, R, C, L, and E can be used to form 6-letter strings such as CIRCLE or CCIRLE. Using these letters, how many different 6-letter strings can be formed in which the two occurrences of the letter C are separated by at least one other letter?
A. 96 B. 120 C. 144 D. 180 E. 240
PS54110.01
Alternatively, a different approach could be to actually list down the cases C_ C _ _ _ The intermediate letters can be arranged in 4! ways. C need not come in the first position and can be moved around 4 spaces to the right(keeping space between the C as one) so 4! * 4 C_ _C _ _ 4! * 3 (C's can be shifted only 3 places to the right) C_ _ _C _ 4! * 2 C_ _ _ _C 4! * 1
Re: The letters C, I, R, C, L, and E can be used to form 6-letter strings
[#permalink]
Show Tags
07 Dec 2019, 17:06
Hi All,
We're told the letters C, I, R, C, L, and E are to be used to form 6-letter strings such as CIRCLE or CCIRLE. We're asked for the number of different 6-letter strings that can be formed in which the two occurrences of the letter C are separated by AT LEAST one other letter. There are a couple of different ways to approach this question - and if you don't know an elegant way to approach it, then you can still get the correct answer with a little permutation math and some 'brute force.'
If the first letter is a C, then the second letter CANNOT be an C (that second letter would have to be one of the other 4 non-C letters)...
C 4
From here, any of the four remaining letters can be in the 3rd spot. After placing one, any of the remaining three letters can be in the 4th spot, etc. and the last letter would be in the 6th spot...
C 4 4 3 2 1
This would give us (4)(4)(3)(2)(1) = 96 possible arrangements with a C in the 1st spot.
If a non-C is in the 1st spot and a C is in the 2nd spot, then we have...
4 C _ _ _ _
A non-C would have to be in the 3rd spot (3 options), then any of the remaining three letters could be 4th, etc...
4 C 3 3 2 1
This would give us (4)(3)(3)(2)(1) = 72 possible arrangements
Next, we could have two non-Cs to start off, then Cs in following spots...
4 3 C 2 1 C --> (4)(3)(2)(1) = 24 possible arrangements 4 3 C 2 C 1 --> (4)(3)(2)(1) = 24 possible arrangements 4 3 2 C 1 C --> (4)(3)(2)(1) = 24 possible arrangements
This would give us an additional (3)(24) =72 possible arrangements
There are no other options to account for, so we have 96+72+72 = 240 total arrangements.