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 digits in the thousand's place and the ten's place are each larger than their neighboring digits.  

 Dec 10, 2017

If we exclude "0" as a beginning digit, we have 4 * 4!  = 96 possible numbers


The "6"  and "8"   in either the tens or thousands positions will each be larger than their neighboring digits.....


And they can be arranged in two ways.....


And given that the 6 and 8  occupy these positions....the first number can be chosen in two ways and the remaining digits can be arranged in 2 ways


So.......2 * 2 * 2  = 8 numbers are possible


So.....the probability is    8/96   =  1/12



cool cool cool

 Dec 10, 2017

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