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07 Sep 2010, 08:47
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
24% (02:22) correct
76%
(01:55)
wrong
based on 72
sessions
History
Date
Time
Result
Not Attempted Yet
How many even numbers greater than 300 can be formed with digits 1,2,3,4,5 no digit being repeated ?
A. 15
B. 48
C. 63
D. 96
E. 111
A. 15
B. 48
C. 63
D. 96
E. 111
Answer: 111
07 Sep 2010, 09:25
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intellijat
Not a GMAT question.
Number can be 3, 4, or 5 digit (4 and 5 digit number will obviously be more than 300).
5 digit even numbers:
Last digit must be either 2 or 4 (for number to be even), so 2 choices. First 4 digits can be arranged in 4! # of ways. So total # of 5 digit even numbers possible is \(4!*2=48\);
4 digit even numbers:
Last digit again must be either 2 or 4, so 2 choices. # of ways to choose rest 3 digits out of 4 digits left when order of the digits matters is \(P^3_4\). So total # of 4 digit even numbers possible is \(P^3_4*2=48\);
3 digit even numbers:
Last digit again must be either 2 or 4, so 2 choices. # of ways to choose rest 2 digits out of 4 digits left when order of the digits matters is \(P^2_4\). So total # of 3 digit even numbers possible is \(P^2_4*2=24\).
But out of these 24 3-digit even numbers some will be less than 300, the ones starting with 1 or 2. There are 9 such numbers (1X2 and 3 options for the second digit X, so 3 plus 1X4 and 3 options for X, so 3 plus 2X4 and 3 options for X, so 3), which means that # of 3 digit even numbers more than 300 is \(24-9=15\);
So total \(48+48+15=111\).
General Discussion
08 Sep 2010, 07:31
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<...> indicates all the choices for each digit
For 3-digit number,
<3, 4, 5> <1, 2, 3, 4, 5> <2, 4>
If 2 is the last digit, <3, 4, 5> <1, 3, 4, 5 minus 1 chosen for first digit> <2> therefore potential combos = (3)(4-1)(1) = 9
If 4 is the last digit, <3, 5> <1, 2, 3, 5 minus 1 chosen for first digit> <4> therefore potential combos = (2)(4-1)(1) = 6
For 4-digit number,
<1, 2, 3, 4, 5 minus 1 chosen for last digit> <1, 2, 3, 4, 5 minus 2 chosen for first&last digit> <1, 2, 3, 4, 5 minus 3 chosen for first&2nd&last digit> <2, 4> therefor potential combos = (5-1)(5-2)(5-3)(2) = 48
For 5-digit number,
<1, 2, 3, 4, 5 minus 1 chosen for last digit> <1, 2, 3, 4, 5 minus 2 chosen for first&last digit> <1, 2, 3, 4, 5 minus 3 chosen for first&2nd&last digit> <1, 2, 3, 4, 5 minus 4 chosen for first&2nd&3rd&last digit> <2, 4> therefor potential combos = (5-1)(5-2)(5-3)(5-4)(2) = 48
Total combos = 9 + 6 + 48 + 48 = 111
For 3-digit number,
<3, 4, 5> <1, 2, 3, 4, 5> <2, 4>
If 2 is the last digit, <3, 4, 5> <1, 3, 4, 5 minus 1 chosen for first digit> <2> therefore potential combos = (3)(4-1)(1) = 9
If 4 is the last digit, <3, 5> <1, 2, 3, 5 minus 1 chosen for first digit> <4> therefore potential combos = (2)(4-1)(1) = 6
For 4-digit number,
<1, 2, 3, 4, 5 minus 1 chosen for last digit> <1, 2, 3, 4, 5 minus 2 chosen for first&last digit> <1, 2, 3, 4, 5 minus 3 chosen for first&2nd&last digit> <2, 4> therefor potential combos = (5-1)(5-2)(5-3)(2) = 48
For 5-digit number,
<1, 2, 3, 4, 5 minus 1 chosen for last digit> <1, 2, 3, 4, 5 minus 2 chosen for first&last digit> <1, 2, 3, 4, 5 minus 3 chosen for first&2nd&last digit> <1, 2, 3, 4, 5 minus 4 chosen for first&2nd&3rd&last digit> <2, 4> therefor potential combos = (5-1)(5-2)(5-3)(5-4)(2) = 48
Total combos = 9 + 6 + 48 + 48 = 111
