The digits 2,4,6,8 and 0 are used to make five digit numbers with no digit repeated. What is the probability that a number chosen at random for these numbers has the property that the digit in the hundreds place is larger than its neighboring digits.
Here is my attempt at your question:
2, 4, 6, 8, 0 = 5 integers. Ordinarily, you would simply have 5! = 120 permutations and each permutation will begin with one of the 5 digits this many times: 120 / 5 = 24 permutations for each digit. But, since you cannot have a 5-digit number that begins with zero "0", then we must subtract 4! = 24 from 120 permutations: 120 - 24 = 96 such 5-digit integers.
For the hundreds' digit to be larger than its neighboring digits, to its left and right, you would have to have the following scenarios:
8=0, 2, 4, 6=4![8 in hundreds' place would have to be surrounded by 0, 2, 4, 6 =4! = 24 ways. The same applies to 6 and 4 as below].
6 =0, 2, 4 =3!
4 =0, 2 =2!
So:[4! + 3! + 2!] =32 such numbers
So, the probability of randomly picking one of these 32 integers out of a total of 96 permutations is: 32 / 96 =1 / 3
