How many five-digit even numbers can be formed using

Question

How many five-digit even numbers can be formed using  6,3,0,2,  and  1  only once?

(A)  32 
(B)  36 
(C)  54 
(D)  60 
(E)  72

chetan2u · Accepted Answer

You can use the digit only once..

Let the number be ABCDE..

(1) First restriction is that  the number is EVEN , so  E can be any of the three 0, 2 and 6---3 ways, the remaining 4 places can be filled in 4*3*2*1 ways 
So ABCDE can be in 4*3*2*1*3=72 ways..

(2) But , the second restriction is that  it is a 5-digit number, so A cannot be 0 
So,  subtract ways in which A is 0 ..
If A is 0,  E can be any of two - 2 or 6. The remaining can be in 3*2*1 ways 
So, ABCDE can be written in 1*3*2*1*2=12 ways..

Our answer =  (1)-(2)=72-12=60

D

gracie · Answer

5*5*4=100 total numbers
3 out of 5 end in even digit (0,2,6)
3/5*100=60 even numbers
D

hudacse6 · Answer

I think this is not a good way for the novice.

ekeiyoung · Answer

I just did a combination.. 5 x 4 x 3 and got sixty. Although I feel like I’m doing it wrong and got lucky.

Posted from my mobile device

navderm · Answer

So Since the most significant digit (MSD) cannot be 0 and yet the numbers can be even only. I split this in two different parts.

P1 : When the lease significant digit (LSD) is 0:
Then there are 4 * 3 * 2* 1 ways to generate this.

P2: When the LSD is even but not 0:
Then 3(all digits except LSD and 0) * 3 * 2 * 1 * 2 (since it has to be even)
= 3 * 3 * 2 *1 *2

= 24 + 36
= 60

hudacse6 · Answer

P2: When the LSD is even but not 0:
Then 3(all digits except LSD and 0) * 3 * 2 * 1 * 2 (since it has to be even)
= 3 * 3 * 2 *1 *2

but if a number that ending with 0, MUST BE EVEN SO IT WILL BE LOOKALIKE  
Then 3(all digits except LSD and 0) * 3 * 2 * 1 *   3   (since it has to be even)
 navderm

navderm · Answer

Hey. I don't quite understand what you're trying to say. 
What I meant was i divided the problem statement to 2 parts, one when the 0 is at the end and another when 0 is not. 
So, when 0 is not at the end we have 2 options (2, 6) to put at the ones place so that the number is even. Hence it won't be
3 as you mentioned.

If i didn't understand your question right, please elaborate and I'll try to explain further.

Archit3110 · Answer

another alternate method to solve
total ways- total odd nos
total ways ; 4*4*3*2*1 ; 96
total odd no; 3*3*2*1*2 ; 36
total even ; 96-36 ; 60
IMO D ;

second method
total even no
case 1 ; last digit is 0 ; 4*3*2*1*1 ; 24
case 2 ; last digit is 2,6 ; 3*3*2*1*2 ; 36
total even ; 24+36 ; 60
IMO D

GMATGuruNY · Answer

Case 1: Units digit is 0
Number of options for the ten-thousands place = 4. (Any of the 4 remaining digits)
Number of options for the thousands place = 3. (Any of the 3 remaining digits)
Number of options for the hundreds place = 2. (Either of the 2 remaining digits)
Number of options for the tens place = 1. (Only 1 digit left)
To combine these options, we multiply:
4*3*2 = 24

Case 2: Units digit is 2 or 6
Number of options for the units place = 2. (2 or 6)
Number of options for the ten-thousands place = 3. (Any of the 3 remaining nonzero digits)
Number of options for the thousands place = 3. (Any of the 3 remaining digits, including 0)
Number of options for the hundreds place = 2. (Either of the 2 remaining digits)
Number of options for the tens place = 1. (Only 1 digit left)
To combine these options, we multiply:
2*3*3*2 = 36

Total ways = Case 1 + Case 2 = 24+36 = 60

D

bumpbot · Answer

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How many five-digit even numbers can be formed using

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