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
i have seen this question on this forum before and i understand that how N could be the first letter but it is hard for me to find out that how 50th word is NAAIG. i know it is in alphabetic order but how it is the 50th word?
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uggghh i am not getting this question
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GMATGuruNY
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Total number of words: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 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.
When the 60 words are listed in alphabetical order, those at the end of the list will all begin with N.
If the first letter is N, the number of ways to arrange the remaining 4 letters AAGI = 4!/2! = 12.
Thus, there are 12 words that begin N.
These 12 words constitute the LAST 12 WORDS in the list:
Words 49 through 60.
Thus:
Word 49 is the FIRST word that starts with N:
NAAGI.
Word 50 is the SECOND word that starts with N:
NAAIG.
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Brent@GMATPrepNow
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I just want to formalize a general principle that Mitch is 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
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
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vipulgoyal
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HI
Check my previous post
https://www.beatthegmat.com/p-c-t202402.html
U can get it by forward counting
words start with A = 24 (1x4x3x2x1)
words start with G = 12 (1x4x3x2x1/2!)
words start with I = 12 (1x4x3x2x1/2!)
so the 49th word in dictonery = NAAGI and 50th will be NAAIG
Check my previous post
https://www.beatthegmat.com/p-c-t202402.html
U can get it by forward counting
words start with A = 24 (1x4x3x2x1)
words start with G = 12 (1x4x3x2x1/2!)
words start with I = 12 (1x4x3x2x1/2!)
so the 49th word in dictonery = NAAGI and 50th will be NAAIG
