How many 3 digits even numbers are possible using the digits 0, 3, 1

Question

Learn when to “Add” and “Multiply” in Permutation & Combination questions- Exercise Question #4

How 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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kunalcvrce · Answer

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

EgmatQuantExpert · Answer

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 3-digit 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

rodrigo.mendozay · Answer

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?

EgmatQuantExpert · Answer

Hey rodrigo.mendozay , Thanks for pointing this in question. We have updated the question

Slomo5000 · Answer

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

antarcticsugar · Answer

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?

Hero8888 · Answer

Hi,
If you set "without repetition" so the answer will change to 36 ways. The guy above is right.

FillFM · Answer

I agree with  Slomo5000  .

I think the answer is 36.

Diwakar003 · Answer

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!

Wildflower · Answer

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!

EgmatQuantExpert · Answer

Hey 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
e-GMAT

EgmatQuantExpert · Answer

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 e-GMAT

EgmatQuantExpert · Answer

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 e-GMAT

Wildflower · Answer

I see where I went wrong. Thank you, EgmatQuantExpert for pointing it out

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How many 3 digits even numbers are possible using the digits 0, 3, 1

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