vipulgoyal wrote:How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

As the integers must be even, the units digit of the integer must be even.
Hence, we have 3 choices (2, 4, 6) for the units digit.

For each of these 3 choices, for other two digits (hundreds and tens), we have 5 and 4 choices.

Hence, total number of such integers = 3*4*5 = 60

Another Method:
Number of 3-digit integers that can be made using the given 6 digits such that no digit is repeated = (Number of ways to select digits for the hundreds place)*(Number of ways to select digits for the tens place)*(Number of ways to select digits for the units place) = (Number of ways to select 1 integer from 6)*(Number of ways to select 1 integer from rest 5)*(Number of ways to select 1 integer from rest 4) = 6*5*4 = 120

As exactly half of the given six digits are even, exactly half of these 120 integers will have even units digit.

Hence, number of even 3-digit integers that can be made using the given 6 digits such that no digit is repeated = 120/2 = 60

vipulgoyal wrote:... if there would be 5 odds and 3 evens and we have to count no of 3 digit even non repeatable numbers then, woulb be "total 3 digit non repeatable divide by 3/8 " right ans

8*7*6 divided by 3/8

Multiplied by NOT divided by.
In the original question the logic is to divide by 6/3, i.e. multiply with 3/6.
Similarly here, multiply with 3/8 or divide by 8/3.

You can cross check the same using the first method also.
We have 3 choices for the units digit and for each of these 3 choices, for other two digits (hundreds and tens), we have 7 and 6 choices.

Hence, total number of such integers = 3*6*7