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05 Oct 2006, 16:57
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3 digit numbers have to be formed with the digits 0,1,2,3,4,5 and no digit is to be repeated in any number. How many even numbers can be formed?
A 48
B 50
C 46
D 52
E 42
A 48
B 50
C 46
D 52
E 42
--== Message from the GMAT Club Team ==--
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THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
05 Oct 2006, 17:26
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hmm... can someone tell me why my logic is wrong?
If it has to be even, then we know that the 3 digit numbers have to end in either a 0, 2, or 4..
The remaining digits on each case can be determined by picking a group of 2 out 5.
Thus, I get 10.
so 10 + 10 + 10 = 30...
but obviously i'm wrong!
If it has to be even, then we know that the 3 digit numbers have to end in either a 0, 2, or 4..
The remaining digits on each case can be determined by picking a group of 2 out 5.
Thus, I get 10.
so 10 + 10 + 10 = 30...
but obviously i'm wrong!
05 Oct 2006, 19:13
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It's 50.
There are 5 possibilities for first digit because it can't be 0.
There are 5 possibilities for the second one.
And there are 2 possibilites for the last one ( we want even number, only 0,2,4 suite us, and one of them was used to for hundreds and tens)
5*5*2=50
There are 5 possibilities for first digit because it can't be 0.
There are 5 possibilities for the second one.
And there are 2 possibilites for the last one ( we want even number, only 0,2,4 suite us, and one of them was used to for hundreds and tens)
5*5*2=50
