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Counting - missing 3's

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Counting - missing 3's

by Brent@GMATPrepNow » Wed Jan 14, 2009 8:32 am
Sid intended to type a seven-digit number, but the two 3’s he meant to type did not show. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?
(A) 10
(B) 16
(C) 21
(D) 24
(E) 27
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by Mr2Bits » Wed Jan 14, 2009 8:46 am
Answer is C) 21

7!/2(5!)

5040/240

or you could just write out all options if your unsure start with the following

33XXXXX
3X3XXXX
etc..

=21
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by Brent@GMATPrepNow » Wed Jan 14, 2009 9:05 am
Answer is C) 21

7!/2(5!)

5040/240
The answer is C (21) - nice work.
Your solution of 7!/2(5!) looks considerably different from mine, but it obviously seems to work. I'd love to hear how you arrived at it.

Here's my approach to the question:

We need to place the two missing 3’s into the number 52115. There are two cases to consider:
(a) The two 3’s are separated by other numbers (e.g., 532135 or 3521135)
(b) the two 3’s appear together (e.g., 5332115 or 5211533)
Case (a): look at the typed number as: _5_2_1_1_5_
The placeholders shown are potential locations to place 3’s. To meet the case (a) criterion of having separated 3’s we need to select 2 of the 6 placeholders and place a 3 in each location. We can do this 6C2 (15) ways
Case (b): Use the same setup: _5_2_1_1_5_
To meet the case (b) criterion of having 3’s together, we need to select 1 of the 6 placeholders and place both 3’s there. We can select 1 of the 6 placeholders 6 ways.
So there are 21 ways (15+6) to place our missing 3’s
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by Ian Stewart » Wed Jan 14, 2009 9:19 am
Mr2Bits' approach works because you can think of the problem as follows:

Say you have seven chairs in a row. There are 21 (7C2) pairs of chairs in which you could place the two 3's. Now fill the five remaining empty chairs in order with the digits 52115. In this way, for each of the 21 different choices we could make for where to place the 3's, we get a different possible seven-digit number, and we clearly get all of them this way.
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by Brent@GMATPrepNow » Wed Jan 14, 2009 11:40 am
Ian Stewart wrote:Mr2Bits' approach works because you can think of the problem as follows:

Say you have seven chairs in a row. There are 21 (7C2) pairs of chairs in which you could place the two 3's. Now fill the five remaining empty chairs in order with the digits 52115. In this way, for each of the 21 different choices we could make for where to place the 3's, we get a different possible seven-digit number, and we clearly get all of them this way.
Very nice! That's what I love about math - lots of ways to reach the same conclusion. This approach is the most eloquent so far!
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