Probability help, plz??

Question

A 6-digit number is created by using each of the digits 3,5,7,9,4 and 8 exactly once. What is the probability that the number will be a multiple of 4 ? Express your answer as a common fraction in lowest term.

The right answer is 1/15 , plz show ALL work-10 pts to best!!

Answer 1

N is 2-digit number divisible by 4.

so N = 48 and 84, giving n(N) = 2

P(multiple of 4)

= n(N) * 4! / 6!

= 2 / 6*5

= 1 / 15

Answer 2

hmm dunno

Answer 3

Ok, so remember that a number is divisible by 4, if the last two digits are divisible by 4.

There are 6! possible numbers. Here are the one divisible by 4:

{975384, 795384, 957384, 597384, 759384, 579384, 973584, 793584, 937584, 397584, 739584, 379584, 953784, 593784, 935784, 395784, 539784, 359784, 753984, 573984, 735984, 375984, 537984, 357984, 975348, 795348, 957348, 597348, 759348, 579348, 973548, 793548, 937548, 397548, 739548, 379548, 953748, 593748, 935748, 395748, 539748, 359748, 753948, 573948, 735948, 375948, 537948, 357948}

There are 48 of them. So the answer is: 48/(6!)= 1/15.