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How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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Updated on: 20 Apr 2018, 22:02
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Learn when to “Add” and “Multiply” in Permutation & Combination questions Exercise Question #4How many 3 digits even numbers are possible using the digits \(0,3,1,6,7,9\) if repetition of digits is allowed? Options:A) 20 B) 30 C) 40 D) 60 E) 120 Learn to use the Keyword Approach in Solving PnC question from the following article: Learn when to “Add” and “Multiply” in Permutation & Combination questions
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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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04 Apr 2018, 07:27
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EgmatQuantExpert wrote: Question: How many 3 digits even numbers are possible using the digits \(0,3,1,6,7,9\)? Options:A) 20 B) 30 C) 40 D) 60 E) 120 let abc is three digit number: a can have 3,1,6,7,9 values so 5 ways. b can have 0,3,1,6,7,9 values so 6 ways. c can have 0,6values so 2 ways. total ways =5*6*2=60 ways D



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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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Updated on: 20 Apr 2018, 22:01
Solution Given:• We are given 6 digits: 0,1,3,6,7, and 9.
To find:• The number of all the 3 digits even numbers from: 0,1,3,6,7, and 9.
Approach and Working:Total ways to make the 3digit number= Ways to fill the hundreds digit AND Ways to fill the tens digit AND ways to fill the units digit Ways to fill the units place: 2 ways • When O is the units digit, OR • When 6 is the units digit Ways to fill the tens place: 6 ways • Tens place can have every digit from0,1,3,6,7, and 9. Ways to fill the hundreds place: 5 ways The important point to note here is that 0 cannot be the hundreds digit.• Hence, hundreds place can have only any digit from 1,3,6,7, and 9.
Hence, total ways= 5*6*2=60 Hence, the correct answer is option D.Answer: D
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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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13 Apr 2018, 07:20
Hi, Is it implicit that if were not told the phrase "pick numbers without repetition", We would assume that we can repeat the numbers for units, tents and hundreds digits?



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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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17 Apr 2018, 07:44
rodrigo.mendozay wrote: Hi, Is it implicit that if were not told the phrase "pick numbers without repetition", We would assume that we can repeat the numbers for units, tents and hundreds digits? Hey rodrigo.mendozay, Thanks for pointing this in question. We have updated the question
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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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17 Apr 2018, 11:31
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EgmatQuantExpert wrote: Solution Given:• We are given 6 digits: 0,1,3,6,7, and 9.
To find:• The number of all the 3 digits even numbers from: 0,1,3,6,7, and 9.
Approach and Working:Total ways to make the 3digit number= Ways to fill the hundreds digit AND Ways to fill the tens digit AND ways to fill the units digit Ways to fill the units place: 2 ways • When O is the units digit, OR • When 6 is the units digit Ways to fill the tens place: 6 ways • Tens place can have every digit from0,1,3,6,7, and 9. Ways to fill the hundreds place: 5 ways The important point to note here is that 0 cannot be the hundreds digit.• Hence, hundreds place can have only any digit from 1,3,6,7, and 9.
Hence, total ways= 5*6*2=60 Hence, the correct answer is option D.Answer: D
Hi, As per my understanding Answer should be 36 Case 1 0 at unit digit 5*4 ways 6 at unit digit 4*4 ways Total 36 Can you tell me where I went wrong?? Sent from my ONEPLUS A5010 using GMAT Club Forum mobile app



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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 07:39
EgmatQuantExpert wrote: Solution Given:• We are given 6 digits: 0,1,3,6,7, and 9.
To find:• The number of all the 3 digits even numbers from: 0,1,3,6,7, and 9.
Approach and Working:Total ways to make the 3digit number= Ways to fill the hundreds digit AND Ways to fill the tens digit AND ways to fill the units digit Ways to fill the units place: 2 ways • When O is the units digit, OR • When 6 is the units digit Ways to fill the tens place: 6 ways • Tens place can have every digit from0,1,3,6,7, and 9. Ways to fill the hundreds place: 5 ways The important point to note here is that 0 cannot be the hundreds digit.• Hence, hundreds place can have only any digit from 1,3,6,7, and 9.
Hence, total ways= 5*6*2=60 Hence, the correct answer is option D.Answer: D
Hi. I am a bit confused. Let's say if 0 or 6 is picked for the unit digit, there should be left with only 5 options on the tens digit and 4 options for the hundreds digit right? since all digits have to be different. can someone please help?



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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 08:39
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EgmatQuantExpert wrote: rodrigo.mendozay wrote: Hi, Is it implicit that if were not told the phrase "pick numbers without repetition", We would assume that we can repeat the numbers for units, tents and hundreds digits? Hey rodrigo.mendozay, Thanks for pointing this in question. We have updated the question Hi, If you set "without repetition" so the answer will change to 36 ways. The guy above is right.



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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 09:01
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Slomo5000 wrote: EgmatQuantExpert wrote: Solution Given:• We are given 6 digits: 0,1,3,6,7, and 9.
To find:• The number of all the 3 digits even numbers from: 0,1,3,6,7, and 9.
Approach and Working:Total ways to make the 3digit number= Ways to fill the hundreds digit AND Ways to fill the tens digit AND ways to fill the units digit Ways to fill the units place: 2 ways • When O is the units digit, OR • When 6 is the units digit Ways to fill the tens place: 6 ways • Tens place can have every digit from0,1,3,6,7, and 9. Ways to fill the hundreds place: 5 ways The important point to note here is that 0 cannot be the hundreds digit.• Hence, hundreds place can have only any digit from 1,3,6,7, and 9.
Hence, total ways= 5*6*2=60 Hence, the correct answer is option D.Answer: D
Hi, As per my understanding Answer should be 36 Case 1 0 at unit digit 5*4 ways 6 at unit digit 4*4 ways Total 36 Can you tell me where I went wrong?? Sent from my ONEPLUS A5010 using GMAT Club Forum mobile appI agree with Slomo5000 . I think the answer is 36.



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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 10:28
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EgmatQuantExpert wrote: Solution Given:• We are given 6 digits: 0,1,3,6,7, and 9.
To find:• The number of all the 3 digits even numbers from: 0,1,3,6,7, and 9.
Approach and Working:Total ways to make the 3digit number= Ways to fill the hundreds digit AND Ways to fill the tens digit AND ways to fill the units digit Ways to fill the units place: 2 ways • When O is the units digit, OR • When 6 is the units digit Ways to fill the tens place: 6 ways • Tens place can have every digit from0,1,3,6,7, and 9. Ways to fill the hundreds place: 5 ways The important point to note here is that 0 cannot be the hundreds digit.• Hence, hundreds place can have only any digit from 1,3,6,7, and 9.
Hence, total ways= 5*6*2=60 Hence, the correct answer is option D.Answer: D
Hello Payal, If the unit's place includes '0', the tenth place cannot include '0' as the question states the numbers cannot repeat. You've considered '0' for both tenth place and ones place in your solution. I used the below approach and I'm getting a different answer, please let me know where I'm going wrong. Last digit is 0 Tens place and hundredth place  5x4 =20 ways. Last digit is 6. Tens placeand hundredth place  4x4 = 16 ways. So a total of 36 ways. Cheers!



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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 10:45
For a 3 digit number to be even, it has to be either 0 or 6. Constraint: all digits have to be differentCase 1: With 0 as the last digit. Units digit = 0 > 1 way Tens digit can be 3,1,6,7,9, and NOT 0 > 5 ways Hundreds digit cannot be 0 and tens digit > 4 ways Total = 1 x 5 x 4 = 20 ways Case 2: With 6 as the last digit Units digit = 6 > 1 way Tens digit cannot be 6 > 5 ways Hundreds digit cannot be units digit, tens digit, and zero > 3 ways Total = 1 x 5 x 3 = 15 ways Total = case 1 + case 2 = 20 + 15 = 35 ways. Could someone please tell me where I went wrong? Thank you!
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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 22:06
Slomo5000 wrote: EgmatQuantExpert wrote: Solution Given:• We are given 6 digits: 0,1,3,6,7, and 9.
To find:• The number of all the 3 digits even numbers from: 0,1,3,6,7, and 9.
Approach and Working:Total ways to make the 3digit number= Ways to fill the hundreds digit AND Ways to fill the tens digit AND ways to fill the units digit Ways to fill the units place: 2 ways • When O is the units digit, OR • When 6 is the units digit Ways to fill the tens place: 6 ways • Tens place can have every digit from0,1,3,6,7, and 9. Ways to fill the hundreds place: 5 ways The important point to note here is that 0 cannot be the hundreds digit.• Hence, hundreds place can have only any digit from 1,3,6,7, and 9.
Hence, total ways= 5*6*2=60 Hence, the correct answer is option D.Answer: D
Hi, As per my understanding Answer should be 36 Case 1 0 at unit digit 5*4 ways 6 at unit digit 4*4 ways Total 36 Can you tell me where I went wrong?? Sent from my ONEPLUS A5010 using GMAT Club Forum mobile appHey Slomo, There is nothing wrong with your method. There was just some confusion while updating the question stem. Apologies for that. If repetition is not allowed then in that case the total cases will definitely be 36. When 0 is at the units place then the tens and hundreds digits can be filled in 5 x 4 ways. However, when 6 is at the units place, then the hundreds place can be filled in only 4 ways and even the tens place can be filled in 4 ways. Thus, total cases possible will be 20 + 16 = 36 ways. Regards, Saquib Quant Expert eGMAT
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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 22:16
Wildflower wrote: For a 3 digit number to be even, it has to be either 0 or 6.
Constraint: all digits have to be different
Case 1: With 0 as the last digit.
Units digit = 0 > 1 way Tens digit can be 3,1,6,7,9, and NOT 0 > 5 ways Hundreds digit cannot be 0 and tens digit > 4 ways Total = 1 x 5 x 4 = 20 ways
Case 2: With 6 as the last digit
Units digit = 6 > 1 way Tens digit cannot be 6 > 5 ways Hundreds digit cannot be units digit, tens digit, and zero > 3 ways Total = 1 x 5 x 3 = 15 ways
Total = case 1 + case 2 = 20 + 15 = 35 ways.
Could someone please tell me where I went wrong? Thank you! Hey Wildflower, I have highlighted the areas where you made a mistake. Whenever you have a situation in which 0 is involved, try to fill that space first, where the confusion will happen (hundreds place in this case)! When 6 has been placed in the units digit, we are left with the following options for tens and hundreds place: 0, 1, 3, 7 and 9 Now, if you fill the tens place first and say that there are five ways to fill it, then you are basically saying that I can put 0 in the hundreds place, I can put 1 also, I can put 3 or 7 or 9 also in the hundreds place. Now think a bit. If you put 6 in the units place and say 0 in the tens place, how many digits are available for the hundreds place? We have 3,7,9 and 1 available right? However, you have written that the hundreds place can be filled in only 3 ways, which is incorrect! So, I hope you understand what cases you are missing out? You are missing those cases, when 0 is put on the tens place. Thus, to avoid such confusion, what we do is that we fill the hundreds place first and say that the hundreds place can be filled in 4 ways (1,3,7 or 9), this ensure that all eligible digits will get a chance to be placed at the hundreds place. Now with are left with 0 and the remaining 3 digits (since one of the digits is already placed at the hundreds place), thus total available cases for the tens place will also be 4 and thus the correct answer will be 4 x 4 = 16. Let me know if you still have any doubts. Regards, Saquib Quant Expert eGMAT
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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 22:18
antarcticsugar wrote: EgmatQuantExpert wrote: Solution Given:• We are given 6 digits: 0,1,3,6,7, and 9.
To find:• The number of all the 3 digits even numbers from: 0,1,3,6,7, and 9.
Approach and Working:Total ways to make the 3digit number= Ways to fill the hundreds digit AND Ways to fill the tens digit AND ways to fill the units digit Ways to fill the units place: 2 ways • When O is the units digit, OR • When 6 is the units digit Ways to fill the tens place: 6 ways • Tens place can have every digit from0,1,3,6,7, and 9. Ways to fill the hundreds place: 5 ways The important point to note here is that 0 cannot be the hundreds digit.• Hence, hundreds place can have only any digit from 1,3,6,7, and 9.
Hence, total ways= 5*6*2=60 Hence, the correct answer is option D.Answer: D
Hi. I am a bit confused. Let's say if 0 or 6 is picked for the unit digit, there should be left with only 5 options on the tens digit and 4 options for the hundreds digit right? since all digits have to be different. can someone please help? Hey, I have updated the question. Please check. If repetition is allowed then 60 cases are possible. If it is not allowed then 36 cases are possible. Regards, Saquib Quant Expert eGMAT
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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1 [#permalink]
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20 Apr 2018, 23:34
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EgmatQuantExpert wrote: Wildflower wrote: For a 3 digit number to be even, it has to be either 0 or 6.
Constraint: all digits have to be different
Case 1: With 0 as the last digit.
Units digit = 0 > 1 way Tens digit can be 3,1,6,7,9, and NOT 0 > 5 ways Hundreds digit cannot be 0 and tens digit > 4 ways Total = 1 x 5 x 4 = 20 ways
Case 2: With 6 as the last digit
Units digit = 6 > 1 way Tens digit cannot be 6 > 5 ways Hundreds digit cannot be units digit, tens digit, and zero > 3 ways Total = 1 x 5 x 3 = 15 ways
Total = case 1 + case 2 = 20 + 15 = 35 ways.
Could someone please tell me where I went wrong? Thank you! Hey Wildflower, I have highlighted the areas where you made a mistake. Whenever you have a situation in which 0 is involved, try to fill that space first, where the confusion will happen (hundreds place in this case)! When 6 has been placed in the units digit, we are left with the following options for tens and hundreds place: 0, 1, 3, 7 and 9 Now, if you fill the tens place first and say that there are five ways to fill it, then you are basically saying that I can put 0 in the hundreds place, I can put 1 also, I can put 3 or 7 or 9 also in the hundreds place. Now think a bit. If you put 6 in the units place and say 0 in the tens place, how many digits are available for the hundreds place? We have 3,7,9 and 1 available right? However, you have written that the hundreds place can be filled in only 3 ways, which is incorrect! So, I hope you understand what cases you are missing out? You are missing those cases, when 0 is put on the tens place. Thus, to avoid such confusion, what we do is that we fill the hundreds place first and say that the hundreds place can be filled in 4 ways (1,3,7 or 9), this ensure that all eligible digits will get a chance to be placed at the hundreds place. Now with are left with 0 and the remaining 3 digits (since one of the digits is already placed at the hundreds place), thus total available cases for the tens place will also be 4 and thus the correct answer will be 4 x 4 = 16. Let me know if you still have any doubts. Regards, Saquib Quant Expert eGMATI see where I went wrong. Thank you, EgmatQuantExpert for pointing it out
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Re: How many 3 digits even numbers are possible using the digits 0, 3, 1
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