3 digit numbers with 0 - 5

100 what u got is right ... but according to me number of odds are 48 and even are 52,

here it goes..
last digit to be odd can be (1 3 5) 3 ways
first digit can be in 4 ways second digit can also be in 4 ways
hence 4*4*3 = 48 ways..

for even

consider last two digits can be 2,4 hence 2 ways..

first digit can be in 4 ways second digit can also be in 4 ways
hence 4*4*2 = 32 ways.

consider zero as its a special case ..

last digit is 0 hence 1 way

first digit can be in 5 ways second digit can also be in 4 ways

hence 5*4*1 = 20 ways

hence number of odds = 48
number of even = 32+20 = 52

total 100..

hope it clear...

Without thinking about it too much you know that there should be a few more evens than odds... It isn't 50/50 because you can't have 0 be the first number... Since you can't have it as the first digit, it becomes slightly more likely for it to be either the second or third digit than any of the other digits. And obviously when 0 is the last digit, the number is even.. so there's going to be a larger number of even numbers than odd numbers. depending on the answer choices... this could be enough to pick the correct answer.

For any non-0 number, the odds that it appears in the 3rd digit is going to be 4/5*4/5*1/4 = 4/25 = 16/100

For 0, it will be: 4/5*1/4 = 1/5 = 5/25 = 20/100

So let's add up all the possible evens:

20/100 + 16/100 + 16/100 (odds of zero, 2 or 4)

and we get 52/100

Can someone explain 5P2?

3 digit number out of 0 1 2 3 4 5
u cant place zero in first place hence u have 5 options
second place u have 5 options as well
third place in 4 ways

hence 5*5*4 = 100 ways

hope it is clear