The letters C, I, R, C, L, and E can be used to form 6-letter strings

Let’s determine the number of ways in which the letters C are next to each other:

[C-C] - I - R - L - E

We treat the two letters [CC] as a single letter since they are always together. Thus, we are in essence arranging 5 letters. This above can be arranged in 5! = 120 ways

The 6 letters C - C - I - R - L - E can be arranged in 6! / 2! = 720/2 = 360.

So, the number of ways to arrange the letters in which the two C’s are separated by at least one letter (i.e., not next to each other) is 360 - 120 = 240.

Answer: E

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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Answer: Option E

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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

Hence E

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


Total=4! * 4 + 4! * 3 + 4! * 2 + 4! * 1= 4!*10 = 240

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

When we "glue" the two C's together, we have a single entity, which means we have a total of 5 different objects to arrange.
We can arrange 5 different objects in 5! ways.
Since there are no duplicates among these 5! possible outcomes, we need not divide by 2

When a letter is repeated 4 times and if we want that there should be at least one letter between each of two occurrences, that case is different from a letter occurring two times and we putting in at least one letter between them.

When there are 2 instances, we glue them together and take them as unacceptable cases. In all other cases, the two are not together.
But when 4 are glued together and we take them as unacceptable and all other cases as acceptable then we are assuming that ...MMMSM..., MMSMIM... etc are acceptable. But that is not what we want.
We don't want even one pair of M's to be together. These questions are handled differently.

Check this:
https://gmatclub.com/forum/find-the-num ... 87580.html

Total Number of possibilities for a 6 digit word = 6! = 720
In the word "Circle" , the letter C is repeated twice so , we divide the value by 2! , we get 6!/2! = 360

Restriction:
The letters C should not be next to each other.
Lets consider both the C as one element , {CC},I,R,L,E - We get total combinations as 5! , which is 120

Final answer
Total possibilities without restriction - number of ways with restriction
360-120 = 240

Ans: E

Can someone answer this question using another approach ?

What I did was arranged the letters C & C keeping a space of atleast one letter between them and arranging the remaining 4 letters in 4! Ways.

That gives me a total of 4! * 4 = 96. Need help !

Posted from GMAT ToolKit

Hi SiddharthR,

Based on your answer, are you assuming that a "C" must occupy the 'first letter" in the word - with the second "C" occupying the third, fourth, fifth or sixth letter? How about when the first letter is NOT a "C"...?

For example: _ C _ _ _ C

Once you consider all of those additional options, you'll have the correct answer.

GMAT assassins aren't born, they're made,
Rich

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

Answer: E


Hi BrentGMATPrepNow, in considering the number of words that DO have adjacent C's, shouldn't the formula be 5!/ 2!? Since the 2 "Cs" are repeated?

Hi BrentGMATPrepNow VeritasKarishma, Please clarify whether my understanding of the above is correct, when we combine/glue similar entities together, they can be treated as unique for example in the word Mississippi

We have 4 S, 4 I, 2 P, and 1 M, if for "Mississippi" we were to find arrangments in which 'S' should be separated by at least one letter and 'I' should also be separated by at least 1 letter then we could treat it as having 4 distinct letters right

1 X, X represents 4 S
1 Y, Y represents 4 I
2 P and 1 M


so the solution to finding arrangements to a problem where all S should be together and so do I's would be below:

Total ways = 11! / (4! * 4! * 2!)

Bad combination (Together) = 4! / 2! (Division by 2! is for repeating P)

Answer = Total ways - Bad combination
Bad combination

Not quite.
If we want arrangements in which 'S' is separated by at least one letter, there are several different ways to break that rule.
- one way to break the rule is to have an arrangement where all 4 S's and all I's are together (e.g., MIIIIPSSSSP)
- another way to break the rule, is to have 3 S's together as in this example: MISSSIPIPSI
And so on

Your solution only addresses the situation where we break the rule by having all 4 S's and all I's are together

Should'nt it be 5!*2! because two c's internally can be placed in either ways?

That would be correct IF the two C's were different.
However, since the two C's are identical, there's only 1 way to arrange them together.

Concepts involved.
Mutually exclusive sets.
Permutation with repetition.