Happy to help!
Now this is tough question. My first question is where did you find this question? It may not be a suitable question for your GMAT prep. My other question is do we have to make actual words from the word "classic" or are we just talking about arranging the letters in the word "classic?" This is an extremely important distinction since the letters in "classic" can be arranged in numerous ways, but if we wanted to arrange the letters in "classic" into other words, we would be limited. Actually, there is only one seven letter word that we can form using the word "classic." The word is "classic."
Please double check this question to make sure that you are quoting it accurately. Otherwise the answer is 1.
virinchiwiwo wrote:
"What fraction of seven lettered words formed using the letters of the words CLASSIC will have the two C's always together?"
I tried using the MISSISSIPPI rule described in
MAGOOSH and got the denominator (7!/(2!*2!)), but I am not able to get the numerator right. Thank you.
Also, I tried solving this question by considering both Cs as single unit, it got me more confused. It'd be great if you you could show how to solve the problem using that method too.
THANK YOU
MAGOOSHRegards,
Sathya
But, I am going to assume that the question is not actually looking for other seven letter words, but merely asking us to rearrange the letters in the word "classic." And I will assume that the question is asking us for unique combinations, not any combination. Let's re-write the problem:
"What fraction of unique seven lettered arrangements using the letters in the word CLASSIC will always have two C's alway together?"
Much better. This is a question we can solve.
The first thing to notice is that we are being asked what is the fraction. That means, I need to first find the total number of combinations and put that in my denominator. Then I will find how many combinations of letters have the two C's next to each other and place that in the numerator. That will be my answer. Let's do the total first.
In the word classic, we have 7 letters. But we can just jump to 7! since we have two repeated letters. Thus we need to use our MISSISSIPPI rule as you indicated:
7! / [(2!)(2!)] =
(7 * 6 * 5 * 4 * 3 * 2 * 1) / (2 * 1 * 2 * 1) =
(7 * 6 * 5 * 4 * 3) / (2 * 1) =
7 * 6 * 5 * 2 * 3 = 1260
Alright so now I know that the total possible unique combinations of the letters in "classic."
So now I have the bottom of my fraction:
x / 1260
Now I need to solve for x, which is the number of combinations in which the C's are always together. At this point, we can use our trick of looking at the C's as a pair. So we can think of the two C's as always together in our 7 letter combinations. Something like this:
C C _ _ _ _ _
_ C C _ _ _ _
_ _ C C _ _ _
_ _ _ C C _ _
_ _ _ _ C C _
_ _ _ _ _ C C
If we think about it this way, then it's like having 6 letters instead of 7. So we can use this to solve. But remember, we still have to account for those two S's. Thus we need to use our MISSISSIPPI rule again:
6! / 2! =
(6 * 5 * 4 * 3 * 2 * 1) / (2 * 1) =
6 * 5 * 4 * 3 = 360
Awesome! Now I know how many unique combinations of the 7 letters in "classic" will have the two C's next to each other. This can go on the top of my fraction and I can reduce:
360 / 1260
36 / 126
6 / 21
2 / 7
And that's our answer!
Does that make sense?
Best of luck with your studies and let me know about the actual wording of this questions, so I can provide a more accurate explanation.
Cheers!
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