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Counting [#permalink] New post 17 Jan 2009, 04:32
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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Re: Counting [#permalink] New post 17 Jan 2009, 09:14
Shelleb17 wrote:
How many words can be formed from the letters of the word ADROIT, which neither begin with T nor end in A ?

Thanks Guys


Here order of letters matters. Therefore, this is a case of permutation.
There are 6 letters. The followings are the positions of 5 letters "ADROIT": 123456
Since T cannot be placed in the first place and A in last place, 5 letters are accounted for rest of the five places.

Here is a little confusion: Whether the repetition of letter is allowed.

1. If repetition allowes: 4x6^5

2. If repetition is not allowed: 6! - 5! - 5! + 4! = 504

6! = total possibilities
5! = T at the first place
5! = A at the last place
4! = Repetation of "T at the first place while A at the last place" or "A at the last place while T at the first place"
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Re: Counting [#permalink] New post 17 Jan 2009, 09:58
if repitition is allowed, then shouldn't the first spot have 5 choices , i.e. all alphabets except 'T' ?
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Re: Counting [#permalink] New post 17 Jan 2009, 10:17
How many words can be formed from the letters of the word ADROIT, which neither begin with T nor end in A ?

number of words formed from the letters are 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
number of words formed with the first letter T 5! = 5 x 4 x 3 x 2 x 1 = 120
number of words formed with the last letter A 5! = 5 x 4 x 3 x 2 x 1 = 120
number of words formed with first letter T and last letter A 4! = 4 x 3 x 2 x 1 = 24

answer is 720 - ( 120 + 120 - 24)
= 504
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Re: Counting [#permalink] New post 17 Jan 2009, 11:56
graduatetutor wrote:
How many words can be formed from the letters of the word ADROIT, which neither begin with T nor end in A ?

number of words formed from the letters are 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
number of words formed with the first letter T 5! = 5 x 4 x 3 x 2 x 1 = 120
number of words formed with the last letter A 5! = 5 x 4 x 3 x 2 x 1 = 120
number of words formed with first letter T and last letter A 4! = 4 x 3 x 2 x 1 = 24
answer is 720 - ( 120 + 120 - 24)
= 504
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number of words formed with first letter T and last letter A 4! = 4 x 3 x 2 x 1 = 24 :roll: - why? Why do you deduct 24?
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Re: Counting [#permalink] New post 17 Jan 2009, 21:07
GMAT TIGER wrote:
number of words formed with first letter T and last letter A 4! = 4 x 3 x 2 x 1 = 24 - why? Why do you deduct 24?


This is because,

number of words formed with the first letter T 5! = 5 x 4 x 3 x 2 x 1 = 120 (This contains the words that are ending with A)
number of words formed with the last letter A 5! = 5 x 4 x 3 x 2 x 1 = 120 (This contains the words that are beginning with T also)

On subtracting these two groups, we are also subtracting twice the group of words beginning with T and ending with A). To eliminate this, we are eliminating one set of these repeating groups.

Hence the solution - 720 - (120 + 120 - 24)
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Re: Counting [#permalink] New post 17 Jan 2009, 22:14
mrsmarthi wrote:
GMAT TIGER wrote:
number of words formed with first letter T and last letter A 4! = 4 x 3 x 2 x 1 = 24 - why? Why do you deduct 24?


This is because,

number of words formed with the first letter T 5! = 5 x 4 x 3 x 2 x 1 = 120 (This contains the words that are ending with A)
number of words formed with the last letter A 5! = 5 x 4 x 3 x 2 x 1 = 120 (This contains the words that are beginning with T also)

On subtracting these two groups, we are also subtracting twice the group of words beginning with T and ending with A). To eliminate this, we are eliminating one set of these repeating groups.

Hence the solution - 720 - (120 + 120 - 24)


Thanks I misread the red part of the question as "neither begin with T nor A"
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Re: Counting   [#permalink] 17 Jan 2009, 22:14
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