Step-by-step explanation:
I assume the digits cannot be repeated in the numbers.
and that we don't want any numbers that start with 0, as that would be considered a 3- digit number, right ?
so, under these assumptions we have the question of in how many ways can we pull 4 digits out of 6.
if the sequence of the pulled digits would not matter, it would be combinations C(6, 4).
but because we are building actual numbers, the sequence does matter, and we use permutations P(6, 4).
P(6, 4) = 6!/(6-4)! = 6!/2! = 6×5×4×3 = 360
so, we can create 360 different numbers.
but not all of them are wanted.
we don't want the ones that start with 0. how many are those ?
since we are handling all digits in the same way and priority, there are an equal amount of numbers that start with a 0, or a 5, or a 6, or a 3, or a 8, or a 7.
so, 1/6 of the 360 numbers start with a 0, and we need to subtract them :
360 - 60 = 300 really 4-digit numbers
and from these 300 we only want even numbers. that means they can only end with 0, 6 or 8.
in the same way as for the first position, we have 1/6 of the more 300 numbers that end with one of these digits.
we allow 3 digits, so we have
3×1/6 × 300 = 1/2 × 300 = 150
that means we can write 150 different 4-digit even numbers out of these 6 digits.
