Posted by SuckatProb on Tuesday, November 20, 2012 at 8:02am.
It means that numbers 2,3,7 have to be first so combination choose 3 out of 5 and then either 4 or 8 can be in the end which can be done in 2! ways. So 20ways out of possible 5!=120 so 20/120= 2/12=1/6
if any number is divisible by 4 , then its last 2 digits must be divisible by 4
e.g 43728 is divisible by 4 because 28 is divisible by 4
whereas 72438 is not , even though its last digit is divisible by 4
obviously it must be also be even,
so the last two digits could be
xxx28
xxx48
xxx24
xxx84
the number of cases for each of these
= 3x2x1
= 6
e.g. for xxx28
34728 , 37428 , 43728 , 47328 , 73428 , 74328
so the number of cases which are divisible by 4 are 3(6) = 24
but the total number of arrangements with no restrictions are 5! = 120
prob of divisible by 4 = 24/120 = 1/5
thanks
none of these answers are correct lol
You are all wrong!!!! Its 3/10
hahaha is RIGHT!!!!!!! ur all worng!!
:) :(
For last 2 digits, you forgot xxx32 and xxx72. There are 4 * 5 = 20 ways to choose the last two digits, and they are all equally likely, so the probability that the number is divisible by 4 is (# of ways to choose last 2 digits such that the number is divisible by 4)/ways to choose last 2 digits = 6/20 = 3/10.
A number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4. Using the given digits, we find that the only two-digit numbers that are divisible by 4 are 24, 28, 32, 48, 72, and 84.
There are $5 \cdot 4 = 20$ ways to choose the last two digits, and they are all equally likely, so the probability that the number is divisible by 4 is $6/20 = \boxed{3/10}$.