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How many even four digit number can be made using all . . .

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by M7MBA » Sat Dec 02, 2017 5:12 am
How many even 4 digit number can be made using all the digits from 2, 4, 5 and 8, without repetition?

A) 4*4*4*4
B) 3*4*4*4
C) 3*3*3*3
D) 3*3*2*1
E) 3*2*2*1

The OA is D.

Experts, how can I get the correct answer here?
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by [email protected] » Sat Dec 02, 2017 11:59 am
Hi M7MBA,

We're asked to find the total number of EVEN 4-digit integers using the digits 2, 4, 5 and 8, but WITHOUT repetition. This question is a permutation question with one specific 'restriction' that we have to deal with first:

Since the 4-digit integer must be EVEN, it MUST end with 2, 4 or 8. There are 3 options for the 4th digit. Once we place one of those digits....
There will be 3 options for the first digit. Once we place one of those digits...
There will be 2 options for the second digit. Once we place one of those digits...
There will be 1 option for the third digit.

Total options = (3)(3)(2)(1) = 18

Final Answer: D

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by Scott@TargetTestPrep » Sun Sep 29, 2019 6:20 pm
M7MBA wrote:How many even 4 digit number can be made using all the digits from 2, 4, 5 and 8, without repetition?

A) 4*4*4*4
B) 3*4*4*4
C) 3*3*3*3
D) 3*3*2*1
E) 3*2*2*1

The OA is D.

Experts, how can I get the correct answer here?

Since the 4-digit number has to be even, the units digit must be either 2, 4 or 8. Therefore, the units digit has 3 choices. Since the digit can't be repeated, the thousands digit now has 3 choices, the hundreds 2 choices and the tens 1 choice. So there are 3 x 2 x 1 x 3 = 3 x 3 x 2 x 1 such 4-digit even numbers.

Answer: D

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