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Find the number of words found by permuting all the letters

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Find the number of words found by permuting all the letters [#permalink] New post 27 Mar 2013, 11:41
00:00
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Difficulty:

  65% (hard)

Question Stats:

52% (03:23) correct 48% (01:49) wrong based on 33 sessions
I'm having issues solving problems like the following two (any help would be greatly appreciated):

1. Find the number of words found by permuting all the letters in the word INDEPENDENCE such that the E's do not come together.

A. 24300 B. 1632960 C. 1663200 D. 30240 E. 12530
[Reveal] Spoiler: OA
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Re: Find the number of words found by permuting all the letters [#permalink] New post 28 Mar 2013, 00:15
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jlo0909 wrote:
I'm having issues solving problems like the following two (any help would be greatly appreciated):

1. Find the number of words found by permuting all the letters in the word INDEPENDENCE such that the E's do not come together.

A. 24300 B. 1632960 C. 1663200 D. 30240 E. 12530


Poorly worded question. Does "do not come together" mean just that all 4 cannot be in a row or that you can never have 2 touch at all?

This question is also too calculation-intensive for the GMAT. A simpler question that doesn't require as much multiplication would make more sense. I will put one at the end of this post. On to this question:

I will assume that Kaplan's wording just means that all four cannot be in a row and that 2 or 3 could be in a row. Besides giving the correct answer, this makes the question easier and more in line with the difficulty of the GMAT. If you had to also calculate out the 2 and 3 in a row, this would be over twice as calculation intensive.

There are two ways to solve these problems, 1) adding up the options that work and 2) subtracting the ones that don't work from the total. I prefer to subtract, but only because it makes more sense to me, so I will do/explain that here.

First, determine the total number of outcomes, including the bad ones (we will subtract the bad ones later). Here we have 12 characters, so there are 12 choices being made. The first choice has 12 options, the second has 11 options ... and so on. When we multiply this, we get 12!.

However, not all of these should be counted because this considers each E, N, and D to be a separate entity. Is FOOD different if we switch the Os and get FOOD? No, so we shouldn't count these as separate outcomes. Instead, we have to divide out these duplicates. Since there are 4 Es, we have to divide our total number by 24 (there are 4*3*2*1 ways to arrange 4 Es). Since there are 3 Ns, we have to divide our total number by 6 (there are 3*2*1 ways to arrange 3 Ns). Since there are 2 Ds, we have to divide our total number by 2. This gives us:

\frac{(12*11*10*9*8*7*6*5*4*3*2*1)}{(4*3*2*1*3*2*2)}

This is basically one big combination formula and it results in 1663200 (don't forget to reduce the fraction on the real GMAT before multiplying - in fact, don't multiply at all until the end so that you can factor some things out). Now, we know that this is too big, so the correct answer must be smaller (we can estimate and guess at this point if time is an issue and if some answers are bigger than this).

To reduce this, we need to subtract the bad outcomes. What would be a bad outcome? EEEEINDPNDNC or EEEEIPCNNNDD or IEEEENDPNDNC

What we see here is that we can put the 4 Es together, but then we have many options for the remaining numbers. We also have more than one place to put the Es. To calculate this, we have 9 places we can put the Es (with 0 other characters before all the way to 8 other characters before) and then 8 other items that need to be arranged (8! outcomes). So multiplying these together to account for 8! outcomes for each of the 9 places gives us 9!.

But then we also have to realize that the Es can be rearranged themselves. They can show up in 4! ways, so we have to multiply the 9! outcomes by this as well. (This will divide out later, but thinking about it is a good habit because some questions, like the one at the end of this post, will give you different items that need to be rearranged).

However, we still aren't done. We still have to divide out the duplicates like we did with the overall number. This gives us:

\frac{(9*8*7*6*5*4*3*2*1)(4*3*2*1)}{(4*3*2*1*3*2*2)}

When we subtract these bad options from the original, we get:

\frac{(12*11*10*9*8*7*6*5*4*3*2*1)}{(4*3*2*1*3*2*2)} - \frac{(9*8*7*6*5*4*3*2*1)(4*3*2*1)}{(4*3*2*1*3*2*2)} = \frac{(9*8*7*6*5*4*3*2*1)}{(4*3*2*1*3*2*2)}*(12*11*10 - 4*3*2*1) = 1260 * 1296 = 1632960

(Again, some of these steps seemed redundant, but sometimes you cannot just simplify it out of the analysis.)

So, the basic steps are:

1. Understand the problem. What can be reordered? What cannot touch?
2. Calculate the total number of outcomes (good and bad). If n is the number of items, this is \frac{n!}{ways.to.arrange.duplicates}
3. Calculate the number of bad options. If n is the number of items and r is the number of things that cannot be together, this is \frac{(n-r+1)!*(r!)}{ways.to.arrange.duplicates}
4. Subtract the bad outcomes from the total outcomes.

The other method (adding up the options that work) is also possible, but it is easier with a smaller number of characters that cannot touch.

A practice questions that i just made up:

1.) How many possible ways can the letters in the word BANANA be rearranged so that the Ns are not touching?

A.) 720
B.) 120
C.) 60
D.) 40
E.) 20

[Reveal] Spoiler: OA
D because \frac{6!}{3*2*2} - \frac{5!(2)}{3*2*2} = 40
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Re: Find the number of words found by permuting all the letters [#permalink] New post 01 Apr 2013, 13:27
mmagyar wrote:


A practice questions that i just made up:

1.) How many possible ways can the letters in the word BANANA be rearranged so that the Ns are not touching?

A.) 720
B.) 120
C.) 60
D.) 40
E.) 20

[Reveal] Spoiler: OA
D because \frac{6!}{3*2*2} - \frac{5!(2)}{3*2*2} = 40


A much better example, much closer to GMAT type.
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Re: Find the number of words found by permuting all the letters [#permalink] New post 04 Apr 2013, 06:49
jlo0909 wrote:
I'm having issues solving problems like the following two (any help would be greatly appreciated):

1. Find the number of words found by permuting all the letters in the word INDEPENDENCE such that the E's do not come together.

A. 24300 B. 1632960 C. 1663200 D. 30240 E. 12530


Ans => total no. of words - no. of words when all E's are together.

Total no. of words = 12!/(3!2!4!) = 1663200

no. of words when all E's are together = consider all 4 E's as a group hence as a single alphabet
= 9!/(3!2!) = 30240

hence, ans = 1663200 - 30240 = 1632960

option B

--------------------hopes this will help ---------------------------------
Re: Find the number of words found by permuting all the letters   [#permalink] 04 Apr 2013, 06:49
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