Probability on having two consecutives letters C on word..

[Coelhothales]

Hi all,

Here is my problem :
Whats the probability on having two consecutives letters C on word CLASSIC
That´s what I thought:

My total possibilities are : 7!
If I want consecutives C´s I would need something like : CC*5! , but there are 6 ways on writing these CC´s. Then my result should be like CC*5!*6 , or just 5!*6

Resolving the probability equation we would have : (5!*6)/7! = 1/7

Am I correct ? Thanks for the support,

Att,

[GMATGuruNY]

Whats the probability on having two consecutives letters C on word CLASSIC

P = (good arrangements)/(all possible arrangements).

All possible arrangements:
There are 7 letters in the word CLASSIC.
The number of ways to arrange 7 distinct elements = 7!.
But CLASSIC is not composed of 7 distinct letters: there are two identical C's and two identical S's.
When an arrangement includes identical elements, we must divide by the number 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 and by another 2! to account for the two identical S's:
7!/(2!2!) = 7*6*5*3*2.

Good arrangements:
In a good arrangement, the two C's are in adjacent positions.
To ensure that the two C's are in adjacent positions, consider CC a single element in the arrangement.
Now count the number of ways to arrange the 6 elements CC, L, A, S, S, and I.
The number of ways to arrange 6 distinct elements = 6!.
To account for the two identical S's, divide by 2!:
6!/2! = 6*5*4*3.

P(the two C's are in adjacent positions) = (6*5*4*3)/(7*6*5*3*2) = 2/7.

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Hi Coelhothales,

Here's another way to approach this question, one that focuses on the probability of the individual elements.

We're given 7 letters that we must work with, we're asked for the probability that the two Cs are next to one another.

I'll start with 7 spaces:

- - - - - - -

The probability that the first 2 spaces are Cs is...

(2/7)(1/6) - - - - - = 2/42

The probability that the second & third spaces are Cs is the same...

- (2/7)(1/6) - - - - = 2/42

The probability that any two consecutive spaces are Cs is the SAME....

So we have....

1st & 2nd = 2/42
2nd & 3rd = 2/42
3rd & 4th = 2/42
4th & 5th = 2/42
5th & 6th = 2/42
6th & 7th = 2/42

Total probability = 12/42 = 2/7

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

[GMATGuruNY]

One more approach.

There are a total of 7 positions in the arrangement:
_ _ _ _ _ _ _

The two C's must occupy 2 of these 7 positions.
Number of ways to choose 2 positions from 7 options = 7C2 = (7*6)/(2*1) = 21.

Among these 21 options, there are a total of 6 ways to place the two C's in adjacent slots:
C C _ _ _ _ _
_ C C _ _ _ _
_ _ C C _ _ _
_ _ _ C C _ _
_ _ _ _ C C _
_ _ _ _ _ C C

Thus:
P(the two C's are adjacent) = 6/21 = 2/7.