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How many even numbers greater than 300 can be formed with
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07 Sep 2010, 08:47
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63% (01:31) correct 38% (01:00) wrong based on 24 sessions
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How many even numbers greater than 300 can be formed with digits 1,2,3,4,5 no digit being repeated ?
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Re: how many even numbers ?
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07 Sep 2010, 09:25
intellijat wrote: how many even numbers greater than 300 can be formed with digits 1,2,3,4,5 no digit being repeated ?
Ans:111 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 3digit 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 \(249=15\); So total \(48+48+15=111\).
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Re: how many even numbers ?
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08 Sep 2010, 07:31
<...> indicates all the choices for each digit
For 3digit 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)(41)(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)(41)(1) = 6
For 4digit 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 = (51)(52)(53)(2) = 48
For 5digit 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 = (51)(52)(53)(54)(2) = 48
Total combos = 9 + 6 + 48 + 48 = 111



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How many even numbers greater than 300 can be formed with
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01 Jun 2016, 08:48
Bunuel wrote: But out of these 24 3digit 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 \(249=15\);
Hello Bunuel, thank you for a great explanation. Would you perhaps mind explaining this part in a bit more detail? Thank you!
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Re: How many even numbers greater than 300 can be formed with
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26 May 2017, 21:45
intellijat wrote: How many even numbers greater than 300 can be formed with digits 1,2,3,4,5 no digit being repeated ? 1. For 3 digits, the hundreds digit can be (3,4,5) and units can be (2,4). There are 5 valid pairs for units and hundreds digits and there can be 3 tens digits for each for total of 15. 2. For 4 digits, the thousands digit can be (1,2,3,4,5) and the units digit can be (2,4). There are 8 valid pairs for the units and thousands digits. the hundreds and the tens digits can be formed in 3P2 = 6 ways for a total of 8*6=48 3. For 5 digits, the ten thousands digit can be (1,2,3,4,5) and the units digits can be (2, 4 ). There are 8 valid pairs for the units and the ten thousands digits and the thousands, hundreds and tens digits can be formed in 3P3=6 ways for a total of 8*6=48 ways 4.Total number of possibilities = (1) + (2) +(3)= 111
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Re: How many even numbers greater than 300 can be formed with
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