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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.  

Guest Dec 10, 2017
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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

CPhill  Dec 10, 2017

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