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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
Correct answer is option 'A'. Can you explain this answer?
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What fraction of seven lettered words formed using the letters of the ...
- This is a typical permutation-probability problem. To make this problem easily understandable we will break into two parts:
i. First, we will find out all the seven lettered words from the letters of word CLASSIC
ii. Next, we will find out how many of these words will have the two C's together. - The total number of words formed using the seven letters from the word CLASSIC is found by using the multiplication principle. There are seven places for each of the seven letters. The first place has 7 choices, the second place has (7-1) =6 choices, the third places has 5 choices and the seventh place has 1 choice. Hence, the total number of words formed is:
= 7 x 6 x 5 x 4 x ... x 1 = 7!
Notice that there are two C's and two S' in the word, which can be treated as repeated elements. To adjust for the repeated elements we will divide 7! by the product of 2! x 2!
So, the total number of words formed is:
7!/(2! x 2!) - We need to find how many of these words will have the two C's together. To do this, let us treat the two C's as a single entity. So, now we have six spaces to fill. Continuing the same way as in the step above, we can fill the first place in 6 ways, the second place in 5 ways and the sixth place in 1 way. Hence there are 6! ways of forming the words. Once again, we will need to adjust for the two S' which can be done by dividing 6! by 2!.
Total number of 7 lettered words such that the two C's are always together = 6!/2! - The fraction of seven lettered words such that the two C's are always together is:
= (number of words with two C's together/total number of words) = (6!/2!)/(7!/[2! x 2!]) = (2!/7) = 2/7 - Hence the correct answer choice is A.
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What fraction of seven lettered words formed using the letters of the ...
- This is a typical permutation-probability problem. To make this problem easily understandable we will break into two parts:
i. First, we will find out all the seven lettered words from the letters of word CLASSIC
ii. Next, we will find out how many of these words will have the two C's together. - The total number of words formed using the seven letters from the word CLASSIC is found by using the multiplication principle. There are seven places for each of the seven letters. The first place has 7 choices, the second place has (7-1) =6 choices, the third places has 5 choices and the seventh place has 1 choice. Hence, the total number of words formed is:
= 7 x 6 x 5 x 4 x ... x 1 = 7!
Notice that there are two C's and two S' in the word, which can be treated as repeated elements. To adjust for the repeated elements we will divide 7! by the product of 2! x 2!
So, the total number of words formed is:
7!/(2! x 2!) - We need to find how many of these words will have the two C's together. To do this, let us treat the two C's as a single entity. So, now we have six spaces to fill. Continuing the same way as in the step above, we can fill the first place in 6 ways, the second place in 5 ways and the sixth place in 1 way. Hence there are 6! ways of forming the words. Once again, we will need to adjust for the two S' which can be done by dividing 6! by 2!.
Total number of 7 lettered words such that the two C's are always together = 6!/2! - The fraction of seven lettered words such that the two C's are always together is:
= (number of words with two C's together/total number of words) = (6!/2!)/(7!/[2! x 2!]) = (2!/7) = 2/7 - Hence the correct answer choice is A.
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What fraction of seven lettered words formed using the letters of the words CLASSIC will have the two Cs always together?a)2/7b)5/7c)15/19d)4/19e)2/8Correct answer is option 'A'. Can you explain this answer?
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