Question 1172438
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Edit: I'm redoing nearly the entire problem. Thanks ikleyn for pointing out I missed that penny. I'm not sure how I overlooked it.


Here's one way to approach the problem. We have 1+3+4+3 = 11 coins total. There are 11! = 39,916,800 ways to arrange them. The exclamation mark indicates factorial. We start at 11, and count our way down to 1, multiplying everything.


So,
11! = 11*10*9*8*7*6*5*4*3*2*1 = 39,916,800
This would be the answer if we could tell the nickels apart from other nickels, the dimes apart from the other dimes, and the quarters from the other quarters.


However, we cannot make such distinctions.


Since we cannot tell the nickels apart, this means we overcounted by a factor of 3! = 6 just for the nickels portion. Similarly, the dimes are overcounted by a factor of 4! = 24 as well. The quarters were overcounted by a factor of 3! = 6.


Overall, we overcounted by a factor of 6*24*6 = 864
We need to divide the result of the 11! by 864 to correct for this erroneous overcount.


Therefore, there are (11!)/(3!*4!*3!) = (39,916,800)/(864) = 46,200 distinct permutations possible.


<font color=red>Answer: 46,200</font></font>