Case1: the 6 in in thousand place and 8 in the tens place, leaving the 0,2,4
_6_8_ -----> 2x2x1 = 4
Case2:
_8_6_ -----> 2x2x1 = 4
Case3:
_8_4_ , the 6 must be the lead digit
68_4_ ------> 2x1 = 2
a total of 10
prob(your event) = 10/96 = 5/48
it would be simple to actually list the numbers
The digits 2, 4, 6, 8 and 0 are used to make five-digit numbers with no digits repeated. What is the probability that a number chosen at random from these numbers has the property that the digits in the thousands place and ten's place are each larger than their neighboring digits.
a)5/48
b)11/96
c)1/8
d)13/96
e)7/48
So I know that it's out of 4 * 4! combinations because you can't have the zero as the first digit, how do I find how many numbers meet the condition?
1 answer