You have 4 quarters, 3 dimes, 2 nickels, and 5 pennies. In how many ways can they be arranged in a row? (Coins of the same value cannot be distinguished from each other.)

Question

anonymous · Answer

I am assuming I just multiply 4*3*2*5?

anonymous · Answer

How are you getting that?

anonymous · Answer

Let $n$ be the total number of coins.  Let $k_i$ be the number of coins in the $i$th denomination.

anonymous · Answer

well because im trying to find out how many ways they can be arranged so i figured if I multiply them i'll get the amount of ways it can be arranged in a row

anonymous · Answer

Just to be clear$$
n = \sum k_i
$$

anonymous · Answer

The number of ways to arrange it would be $$
\frac{n!}{\prod k_i!}
$$

anonymous · Answer

This is a generalized version of the choose operation.  The choose operation is basically $k_1=k$ and $k_2=n-k$. $$
\frac{n!}{k_1!k_2!} = \frac{n!}{k!(n-k)!}
$$

anonymous · Answer

In our case, we have: $$
\frac{(4+3+2+5)!}{4!3!2!5!}
$$

anonymous · Answer

Which is a very big number it seems... http://www.wolframalpha.com/input/?i=(4%2B3%2B2%2B5)!%2F(4!3!2!5!)

anonymous · Answer

Pretty interesting pattern when you think about it: $$\Large
\frac{\left(\sum k_i\right)!}{\prod \left(k_i!\right)}
$$