How many 4-digit numbers can be made that are divisible by 4 if we have the digits 2, 4, 5, 6, 7 and for each number we can only use each digit once?

1 Answer

36

Explanation:

The first thing to know is that any number is divisible by 4 if the last two digits of that number are divisible by 4.

http://www.aaamath.com/fra72_x5.htm

So the first question is what 2 digit numbers, with no digits repeating, that are multiples of 4, can we make with the digits?

#24, 52, 56, 64, 72, 76# - and so there are 6 numbers.

With those 6 numbers now known, we can fill in the other two digits of our number. We'll be taking the remaining 3 numbers from our group of 5 (so when the last two digits are 2 and 4, we can use the 5, 6, and 7) and putting them into the first two spots in our number. We do care about order, and so we'll use the permutation general formula:

#P_(n,k)=(n!)/((n-k)!); n="population", k="picks"#

#P_(3,2)=(3!)/((3-2)!)=(3!)/(1!)=3! = 6#

All told, we have 6 ways to order the two final numbers and 6 ways to order the first two numbers, giving:

#6xx6=36#