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Permutations and Combination

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by regor60 » Tue Jan 02, 2018 9:26 am
Roland2rule wrote:What fraction of seven lettered words formed using the letters of the words CLASSIC will have the two C's always together?

A) 2/7
B) 5/7
C) 15/19
D) 4/19
E) 2/8

OA is A
Is there a unique formula to solve this question? An Expert answer is needed here.
First determine how many 7 letter "words" can be formed with these 7 letters.

The first letter can be any of the 7. The next any of the remaining 6 and so on. So 7x6x5x...1 = 7!

Recognize that two of the letters, C, are the same, so need to divide by 2.

So the total possible 7 letter words are 7!/2. That is the denominator of the fraction.

Now, determine how many of these have 2 C's together. Start with the two CCs in positions one and two. The remaining 5 letters can be arranged as discussed above 5x4x3x...1 = 5!

Now, march the two CC's through the letter positions and you can see there are 6 in total. So the total number of words with CC's together is 6x5! = 6!

Divided 6! by 7!/2 =6!*2/7! = [spoiler] 2/7, A[/spoiler]
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by Scott@TargetTestPrep » Wed Aug 14, 2019 4:31 pm
BTGmoderatorRO wrote:What fraction of seven lettered words formed using the letters of the words CLASSIC will have the two C's always together?

A) 2/7
B) 5/7
C) 15/19
D) 4/19
E) 2/8

OA is A
Is there a unique formula to solve this question? An Expert answer is needed here.
The total number of seven-letter words that can be formed using the letters in CLASSIC (regardless if the 2C's are together or not) is calculated by using the indistinguishable permutations formula, noting that C appears twice and S appears twice in the word CLASSIC:

7!/(2! x 2!) = 5040/(2 x 2) = 1,260.

The total number of seven-letter words that can be formed if the 2 C's are together (by considering the 2C's as one letter) is:

6!/2! = 720/2 = 360

Therefore, the fraction is 360/1,260 = 36/126 = 6/21 = 2/7.

Answer: A

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Founder and CEO
[email protected]

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