What is the average (mean) of all $$5-$$digit numbers that can be formed by using each of the digits $$1$$, $$3$$, $$5$$, $$7$$, and $$8$$ exactly once? ($$2005$$ AMC $$10B$$ Problem, Question \#$$20$$) 
Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Method $$1$$: We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $$4!=24$$ ways to arrange the other numbers, so each number appears in each spot $$24$$ times. Therefore, the sum of all such numbers is $$24\times (1+3+5+7+8)\times (11111)=24\times24\times11111=6399936$$.  Since there are $$5!=120$$ such numbers, we divide $$6399936\div120$$~ to get $$53332.8$$.  Method $$2$$: We can first solve for the mean for the digits $$1$$, $$3$$, $$5$$, $$7$$, and $$9$$ since each is $$2$$ away from each other. The mean of the numbers than can be solved using these digits is $$55555$$. The total amount of numbers that can be formed using these digits is $$5!=120$$. The sum of these numbers is $$55555(120)=6666600$$. Now we can find out the total value that was gained by replacing the $$8$$ with a $$9$$. We can start by calculating the gain when the $$8$$ was in the ones digit. Since there are $$4!=24$$ numbers with the $$8$$ in the ones digit and $$1$$ was gain from each of them, $$24$$ is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $$24+240+2400+24000+240000=266664$$ as the total amount that was gained. Subtract this amount from the sum of the digits using the~ $$9$$ instead of the $$8$$ to get $$6666600-266664=6399936$$. Finally, we divide this by $$120$$ to get the average $$\frac{6399936}{120}=53332.8$$.  Method $$3$$: The average value of the digits is $$\frac{(1 + 3 + 5 + 7 + 8)}{5} = 4.8$$. Values occur in every place so $$4.8 \times 11111 = 53332.8$$. 