Step-by-step explanation:
I guess that means the digits can not be repeated in such a number.
let's start with how many numbers in general can be created :
we have 6 basic digits, and we are pulling 4 of them for a number.
if the sequence of the pulled digits would not matter, it would be combinations C (6, 4).
but we are creating different numbers, so the sequence of the digits does matter. so, it is permutations P (6, 4).
P(6, 4) = 6!/(6-4)! = 6!/2! = 6×5×4×3 = 360
so. bauxite, we can create 360 different 4 digit numbers out of these 6 digits.
but not all of them are wanted.
for example, I assume we don't want the numbers that start with a 0, because they would normally count as 3-digit numbers.
if that assumption is correct, we need to find how many would start with a 0, and subtract those from the total number.
since we are handling all 6 digits in the same way, an equal number must start with 0, with 5, with 6, with 3, with 8 and with 7.
so, 1/6 of the total number start with 0 : 360×1/6 = 60.
that means the total number of 4-digit numbers out of these 6 digits that do not start with 0 is 360-60 = 300.
but we are still not finished : we only want even numbers. that means they end with 0, 6 or 8.
in the same way we considered the first position, we consider also the last position. we have an equal amount of numbers that end with the different digits.
and so, we have 1/6 of 300 for each ending digit.
we want the specified 3 digits as end digits, so we get
3×1/6 of the 300 = 1/2 × 300 = 150
so, we can write 150 different 4-digit even numbers out of these 6 digits.
