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# How many words can be formed from the letters of the word

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How many words can be formed from the letters of the word [#permalink]

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19 Sep 2008, 02:19
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How many words can be formed from the letters of the word ADROIT, which neither begin with T nor end in A ?
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Last edited by amitdgr on 19 Sep 2008, 10:55, edited 1 time in total.

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VP
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19 Sep 2008, 03:08

Thanks
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19 Sep 2008, 10:20
Hi..

It is simple.

so we can 6 letter words with A, D, R, O, I and T without repeatation. Here, we can assume that we are making only 6 letter words. so it is 6!.

Now first can not be T. So any other 5 letters not T. = 5!

similarly, last can not be A. so any other 5 letters not A = 5!.

6! - ( 2 * 5! ) = 6*5! - 2*5! = 4*5! = 480.

But if we have to make 6 letters words with repeatations of letters.

Then I think 5 letter for first, 6 letter for second, third, fourth and five, and again 5 for last letter of the word. if would come = 5*6*6*6*6*5 = 6480.

Please correct me if i am wrong.

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19 Sep 2008, 10:38
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OA is 504. The explanation, of which I understand only a part, is as follows

Case 1: Words that begin with A

Since A is in the first place, it takes care of both the conditions given(that T shouldn't be first place and A shouldn't be the last). Now there are 5! words that begin with A

Case 2: Words that do not begin with A

Here we have to ensure that T does not come in the first place. So the first place can be filled in 4 ways(using any letter except A and T). Out of the remaining 5 letters A cannot go into the last place. Hence last place can be filled in 4 ways. The other 4 places can be filled in 4! ways. Hence the number of words are 4 * 4! * 4

Combining both the cases, the total number of words are 5! + 16 * 4! = 504

I do not understand most part of the approach.
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19 Sep 2008, 10:46
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Alternate approach:

6 different letters. so total no of words possible: 6!

To determine number of words that can be formed which neither begin with T or end with A , we first determine the words which EITHER begin with T OR end with A

There are 5! words which begin with T and 5! words that end with A, while the number of words that begin with T and end with A are 4!. Hence of the 5! + 5! words that begin with T or end with A, we exclude words which begin wit T AND end with A. Hence reqd no of words are 5!+5!-4! = 216

number of words that can be formed which neither begin with T or end with A = total no of words possible - the no. words which can be formed which begin with T or end with A

= 720-216=504
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Re: PS: Weird Word Formation question (Combinatorics) [#permalink]

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20 Sep 2008, 03:48
Do we get such questions on GMAT ?
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Re: PS: Weird Word Formation question (Combinatorics) [#permalink]

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21 Sep 2008, 22:07
I think the wording of the question is playing tricks here. The question stem says:

“neither begin with T nor end in A”.

The solution 6! – 2*5! = 480 EXCLUDES all the words which

“begin with T AND end in A” (this number is 1*4!*1 = 24)

I still think we should be excluding words that “begin with T AND end in A”. What is the source of this question?

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Re: PS: Weird Word Formation question (Combinatorics) [#permalink]

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21 Sep 2008, 22:15
ankush83gupta wrote:
I think the wording of the question is playing tricks here. The question stem says:

“neither begin with T nor end in A”.

The solution 6! – 2*5! = 480 EXCLUDES all the words which

“begin with T AND end in A” (this number is 1*4!*1 = 24)

I still think we should be excluding words that “begin with T AND end in A”. What is the source of this question?

I fell for the trap an I computed the “begin with T AND end in A” and subtracted it from 6!.

On a close look. Neither begin with T nor end in A includes those do not begin with T and not end in A. So we have to account for them as well.

Hence the original solution makes sense.

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Re: PS: Weird Word Formation question (Combinatorics)   [#permalink] 21 Sep 2008, 22:15
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