Answer choices?

But here's my best go at it... probably not the quickest, but hopefully it helps.

102 120 130 140 150
103 123 132 142 152
104 124 134 143 153
105 125 135 145 154

I started out with the number 1 since a 3 digit number can't begin with zero.

When I did the actual problem, I just did the first column 102-105 above (I've expanded it with the remaining 4 columns just to help show my reasoning).

After that I assumed (hopefully, correctly) that there would be 4 more groups of the 4 rows by five columns worth of numbers above.

So, in one group there are 20 numbers. Then, in 5 total groups there would be 100 numbers.

As far as odds and evens go, I'm a little less sure on this rationale, but maybe we apply the same logic?

The group above has 12 evens and 8 odds. Multiply each by five for a total of 60 evens and 40 odds... My reasoning for this portion may be flawed, but this is my best guess at it.

J

for even u can have 0,2,4 at the end
suppose i have 0 at the end then
_ _ 0

now i can fill tens place in 5 ways (12345) and hundreds place in 4 ways .
therefore once u have zero in end we get 5*4 = 20 combinations

Let zero be the the center
_ 0 _
now we can fill units place in two ways (2,4) and hundreds place in 4 ways ..therefore combinations = 8

as we cantuse zero at hundreds place so lets not use zero anymore
now we can fill units place on 2(2,4) ways , tens in 4 , hundreds in 3
so we have 2*4*3 = 24

so ans = 20 + 8 + 24 = 52

crackgmat007 wrote:How many number of 3 digit numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number? How many of these are even?

Pls explain the concept for finding the # of even numbers. Tx.

total number of 3 digit numbers(that includes numbers beginning with 0)=
6C3*3!=120
now we need to find the numbers beginning with 0 and subtract it from 120
numbers are of the form 0XX
so numbers beginning with 0=5C2*2!=20
so we can form 120-20=100 3 digit numbers

NUMBER OF EVEN NUMBERS

in this case numbers must end with 2,4,0
numbers ending with 2=4C2*2!=12
numbers ending with 4=12
numbers ending with 0=5C2*2!=20
total =44

100 is correct.

First assume all three digit numbers formed by 6 numbers = 6P3 or 6!/3!.

Next for the zeros; you can have 012, 013, 014, 015, 023, 024, 025, 034, 035, 045. Each of the two numbers following zero can be interchanged so you have 2!. so you have 5C2*2 = 20. So three digit numbers = 120 - 20 = 100

For even it's 52 not 44.

If we have _ _ 2, then we have 4 spots in the first and 4 spots in the second, so we have 16

Same applies if for _ _ 4

For _ _ 0, we have 5 spots in the first and 4 spots in the second. So we have a total of 16 + 16 + 20 = 52

tohellandback wrote:
NUMBER OF EVEN NUMBERS

in this case numbers must end with 2,4,0
numbers ending with 2=4C2*2!=12
numbers ending with 4=12
numbers ending with 0=5C2*2!=20
total =44

could u please elaborate some more on this method..how do u get the numbers ending with 2, 4, or 0..

ketkoag wrote:

tohellandback wrote:
NUMBER OF EVEN NUMBERS

in this case numbers must end with 2,4,0
numbers ending with 2=4C2*2!=12
numbers ending with 4=12
numbers ending with 0=5C2*2!=20
total =44

could u please elaborate some more on this method..how do u get the numbers ending with 2, 4, or 0..

sure..
numbers ending with 2, we don't have to count the numbers that begin with 0. so numbers are of the form XY2 and X should not be 0.
so we need to select 2 numbers from 1,3,4,5 and X and Y can be arranged in 2! ways . in other words they can be interchanged
thats why 4C2*2!- you can call this 4P2 if you wish but I like to do it like this
same goes for numbers ending with 4
numbers ending with 0: we can select any of the two numbers from 1,2,3,4,5
thats why 5C2*2!- again you can call this 5P2