Answer is B.

If we can calculate all the odd integers below 500 with distinct digits and remove them from 250, would be the answer required.

Odd integers starting with 1 can be give as 1AB where A can take 9 values and B can take 4 values (from 1,3,5,7,9 we shd not consider 1 as we are trying to get all distinct digits)
so number of odd integers starting with 1 are 9*4 =36. The same condition holds tru for numbers starting with 3
Similarly numbers starting with 2,4 are 9*5 = 45 (respective places of A and B)

so total number of odd integers below 500 that are distinct are 36*2+ 45*2 = 162.
So the required answer = 250-162 =88

Regards,
Chetan

There are only 200 THREE-DIGIT odd numbers below 500, so we should start with 200, rather than 250.

You are right to consider 1 and 3 separately from 2 and 4.
However, for 1 and 3, there actually
8*4 options, because once you choose the last digit, there are only 8 options left for the tens digit

And, for 2 and 4, there actually
8*5 options, because once you choose the last digit, there are only 8 options left for the tens digit

So the total with distinct is 2*8*4 + 2*8*5 = [spoiler]64 + 80 = 144. And then we subtract 200-144=56. A.[/spoiler]

gmatboost wrote:There are only 200 THREE-DIGIT odd numbers below 500, so we should start with 200, rather than 250.

You are right to consider 1 and 3 separately from 2 and 4.
However, for 1 and 3, there actually
8*4 options, because once you choose the last digit, there are only 8 options left for the tens digit

And, for 2 and 4, there actually
8*5 options, because once you choose the last digit, there are only 8 options left for the tens digit

So the total with distinct is 2*8*4 + 2*8*5 = [spoiler]64 + 80 = 144. And then we subtract 200-144=56. A.[/spoiler]

Oh... what a blunder... how did I miss that!!!
Thanks for correcting me... dude

Regards,
Chetan