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by vipulgoyal » Thu Mar 14, 2013 1:55 am
Find the number of words with or without meaning which can be made
using all the letters of the word AGAIN. If these words are written as in a dictionary,
what will be the 50th word?

Ans NAAIG
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by Anju@Gurome » Thu Mar 14, 2013 2:48 am
vipulgoyal wrote:Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word?
Number of words possible = 5!/2! = 60

As 50 is very close to 60, the total number of possible words, there is fair chance that the 50-th word will start with 'N'. So we will start checking from the end, i.e. we will try to find out the 11th word from the end when the words are arranged alphabetically.

From this point it is easy to list down the last 11 words and find out the word.
However, for a general discussion I'm continuing the discussion to show how to find the word when it is not possible to list down the words.

Now, if the first letter is 'N', number of possible words = 4!/2! = 12
As 12 > 11, the 11-th word from last will definitely start with 'N'.

Now, the 12-th word from the last must be the first word starting with 'N' when all the words starting with 'N' is arranged alphabetically ---> NAAGI

And the 11-th word from the last must be the second word starting with 'N' when all the words starting with 'N' is arranged alphabetically ---> NAAIG
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by Anju@Gurome » Thu Mar 14, 2013 2:58 am
A more general approach is to start calculating the number of words from the start, i.e. in alphabetical order. I'll describe that here...

Number of words starting with 'A' = 4! = 24
As 24 < 50, the word must be starting with some other letter.

Number of words starting with either 'I' or 'G' = 4!/2! = 12
As (24 + 12 + 12) < 50, the word must be starting with 'N'.

Now, (24 + 12 + 12) = 48 words start with either 'A' or 'G' or 'I', the 50-th word must be the second word starting with 'N', which is NAAIG
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by neha24 » Thu Mar 14, 2013 3:10 am
another way !!
the dictionary will follow a sequence as A ,G,I,N in alphabetical order
so we have 5 positions to be filled
case 1 with A in 1st position we will have 4! =24 ways
case 2 : with G in first position we will have 4!/2! =12 ways
case 3 : with I in 1st position we will again have 12 ways
so total we have 48 letters by now so 2 more to go !!
case 4 will start with letter N in first position so 49th word is NAAGI
and hence 50th word is NAAIG

oops i didnt see by the time i typed anju maam had already the other way posted !!
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by GMATGuruNY » Thu Mar 14, 2013 5:51 am
vipulgoyal wrote:Find the number of words with or without meaning which can be made
using all the letters of the word AGAIN. If these words are written as in a dictionary,
what will be the 50th word?
Total number of words:
Number of ways to arrange 5 elements = 5!.
When a permutation includes IDENTICAL elements, we must divide by the number of ways to ARRANGE the identical elements.
The reason is that the permutation doesn't change when the identical elements swap positions.
Number of ways to arrange the two identical A's = 2!.
Thus:
Total possible permutations = 5!/2! = 60.

50th word:
Work BACKWARDS from the END of the list.
If the first letter is N, the number of ways to arrange the remaining 4 letters AAGI = 4!/2! = 12.
Thus, the last 12 words -- words 49 through 60 -- begin with N:
Word 49 = NAAGI.
Word 50 = NAAIG.
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by Brent@GMATPrepNow » Thu Mar 14, 2013 7:14 am
I just want to formalize a general principle that people are using to count the arrangements if the 5 letters in AGAIN.

When we want to arrange a group of items in which some of the items are identical (like having duplicate letters in AGAIN), we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical O's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

In the word AGAIN...
There are 5 letters in total
There are 2 identical A's
So, the total number of possible arrangements = 5!/(2!)
= 60

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by sana.noor » Thu Mar 21, 2013 6:29 am
Is it a gmat style question?
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by Brent@GMATPrepNow » Thu Nov 06, 2014 1:46 pm
sana.noor wrote:Is it a gmat style question?
It's missing the 5 answer choices, so that alone makes in NOT GMAT-like.
If we added 5 answer choices, I think it would be borderline. That is, it would be 750+

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by Mathsbuddy » Wed Nov 12, 2014 6:10 am
Brent@GMATPrepNow wrote:I just want to formalize a general principle that people are using to count the arrangements if the 5 letters in AGAIN.

When we want to arrange a group of items in which some of the items are identical (like having duplicate letters in AGAIN), we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical O's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

In the word AGAIN...
There are 5 letters in total
There are 2 identical A's
So, the total number of possible arrangements = 5!/(2!)
= 60

Cheers,
Brent
Thanks Brent,

I think that one way to understand your MISSISSIPPI rule in the context of the "AGAIN" problem is as follows.
For convenience (but not really necessary)I will start with the letters in alphabetical order:
AAGIN in positions 0,1,2,3 and 4 respectively.
If the first letter "A" is temporarily ignored, then the last 4 letters have 4 x 3 x 2 = 24 arrangements across positions 1,2,3 and 4.
Now the unused "A", which I will call "B", can be squeezed in before 1, between 1 and 2, between 2 and 3, between 3 and 4, or after 4:
Bxxxx
xBxxx
xxBxx
xxxBx
xxxxA
In other words there are 5 ways of positioning the spare "A" (known here as B)

So now we have 24 arrangements x 5 ways = 120 (or 1x2x3x4x5 = 5!) combinations
However, each arrangement appears twice due to the fact that A = B
e.g. ABxxx = BAxxx

Therefore we must halve our combinations: 120/2 = 60

The principle can be extended to any number of replicated letters, creating your MISSISSIPPI formula exactly.

I don't know if this is any help to anyone, but for those who don't like accepting formulae blindly, I hope it gives a little insight.
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by GMATinsight » Fri Nov 28, 2014 11:08 pm
vipulgoyal wrote:Find the number of words with or without meaning which can be made
using all the letters of the word AGAIN. If these words are written as in a dictionary,
what will be the 50th word?

Ans NAAIG
Total Possible outcomes = 5!/2! = 120/2 = 60

Alphabetically:

'A' is fixed at first place then total cases = 4! (for remaining 4 letters) = 24

'G' is fixed at first place then total cases = 4!/2! (for remaining 4 letters with 2 A) = 12

'I' is fixed at first place then total cases = 4!/2! (for remaining 4 letters with 2 A) = 12
Total cases counted = 24+12+12 = 48

49th Case = 'N'AAGI
50th Case = 'N'AGAI
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