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tough one - combinations

Expert replies

by GmatMathPro » Thu Oct 20, 2011 8:42 pm
So when I say S4=4, what I'm saying is that every single unique sequence where we have four matching pairs will be counted 4 times each.

When we do 4C1*7!/2!2!2!, it's counting all sequences where aa is together, all sequences where bb is together, all sequences where cc is together, and all sequences where dd is together.

If we examine a specific sequence, like aabbccdd,

It will get counted when we count how many sequences have aa, it will also get counted when we count how many sequences have bb, cc, and dd. So we're counting this sequence 4 times, but it only should count as one.

But this is true of EVERY sequence like that. bbaaccdd also gets counted four times because it also is counted each time we count the aa's bb's cc's and dd's.

Picture a 4-circle venn diagram. The region that contains all four circles is the set of sequences that have four adjacent matching pairs. So when we add all the ones that have aa, all the ones that have bb, all the ones that have cc, all the ones that have dd, that region is getting counted 4 times, and that is why S4=4.

I'm not 100% sure I understood your question, so I hope this helps.

You are right that aabbccdd is different from bbaaccdd. My point is just that each of these unique sequences is counted four times when we do 4C1*7!/2!2!2!
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by shankar.ashwin » Thu Oct 20, 2011 9:17 pm
Just a thought here; maybe I am completely wrong , but could someone clarify.
I saw the answer for this one and was working backward to see if we could use the slot method here, I got _ _ _ _ (_ _ _ _)

We could fill these up in 4*3*3*3 ways for the first 4 which is understandable.

For for the last 4, I get (2*2*2*1) ways of arranging them.

In total it would be, 4*3*3*3*2*2*2*1=864.

I know the answer is correct here but, I dont get why the 5th and 6th position would have only 2 possibilities here. If someone could try using the slot method here, any idea? For I believe we could do this using the slot method as well. Pls correct me if I am wrong :)
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by GmatMathPro » Thu Oct 20, 2011 10:06 pm
Well, the answer for the same problem with only three letters, that is, how many ways can abcabc be rearranged such that no two adjacent letters are matching is 30.

If slot method works for this type of problem, you'd have to justify that 5*3*2*1*1*1 makes sense somehow.
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by shankar.ashwin » Fri Oct 21, 2011 12:08 am
Ah! I agree that was an absurd question to ask. It doesn't make sense
GmatMathPro wrote:Well, the answer for the same problem with only three letters, that is, how many ways can abcabc be rearranged such that no two adjacent letters are matching is 30.

If slot method works for this type of problem, you'd have to justify that 5*3*2*1*1*1 makes sense somehow.
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by sl750 » Fri Oct 21, 2011 2:09 am
Thanks for clarifying, Pete!.

The solution I posted obviously doesn't work for cases with an odd number of unique letters
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by saketk » Thu Nov 03, 2011 11:04 am
GmatMathPro wrote:Notes:

1. In my opinion this is too hard for a real GMAT question.
Yes, you are absolutely correct. This question is taken from one of India's MBA entrance exam aka CAT (Common admission test). CAT Math level is more difficult than GMAT Math.

Here is the link --

https://www.pagalguy.com/forum/quantitat ... ost3003257
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