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Combinatorics #1

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by papgust » Sun Oct 18, 2009 5:08 am
How many numbers between 200 and 1200 can be formed with the digits 0,1,2,3 if repetition of digits is allowed?

A. 48
B. 63
C. 32
D. 14
E. 20

My Solution:
First, i'm splitting 3-digit and 4-digit sets like 201-999 as set-1 and 1000-1199 as set-2

For set-1, Hundred-digit * tens-digit * units-digit = 2 possibilities * 4 possibilities * 4 possibilities => 32

Similarly for set-2, Thousand-digit * Hundred-digit * tens-digit * units-digit = 1 possibility * 2 possibilities * 4 possibilities * 4 possibilities => 32

Therefore, 32+32 in total = 64.
But the OA is 63. Why do we subtract the total by 1 here? Can someone clarify my doubt and explain? This is not a GMAT problem for sure. I'm very weak in this subject and i'm working hard on it to build my fundamentals. I hope my steps to solve the problem are correct.
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by NikolayZ » Sun Oct 18, 2009 9:05 am
Hey mate.
You included number 200 in your answer, that is why you have to subtract 1 from 64.

3 digit integer.
XXX
Hundreds - 2,3 - total 2
Tens - 0,1,2,3 - total 4
units - 0,1,2,3 - total 4
Look, Number 200 is totally allowed by the set we have established, hence you have to subtract 1.

Hope it helped.
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by crackgmat007 » Sun Oct 18, 2009 12:58 pm
I dont get it. Looks like both 200 & 1200 are included in the computation. Am I missing something?
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by Talkativetree » Sun Oct 18, 2009 5:10 pm
I did this problem by calculating the total number of possible combo's for 1200 and subtracting the total number of possible combo's for 0000-0200.

# of possibilities

0000-1200
2x3x4x4=96

0000-0199(0200)
1x2x4x4 = 32+1

so total possibilities would equal 96-33=63?

either way, I would think 1200 wouldn't be in the set eitehr, so then wouldn't it be 62?
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by papgust » Sun Oct 18, 2009 11:03 pm
Thanks NikolayZ,

That helps and cleared my doubt.
Talkativetree wrote: either way, I would think 1200 wouldn't be in the set eitehr, so then wouldn't it be 62?
Yes, 1200 is not included here. Otherwise it would be 62.
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