A group of 30 dice are thrown. What is the probability that 5 of each of the values 1 , 2 , 3 , 4 , 5 , 6 appear? A group of 30 dice are thrown. What is the probability that 5 of each of the values 1 , 2 , 3 , 4 , 5 , 6 appear?
I solved the problem myself and got (30!)/(5!5!5!5!5!5!) as an answer, since you want 5 of 6 different values. However, when looking at the solutions provided by my professor, it says the answer is (30!)/(5!5!5!5!5!).
 A: Just an alternate approach to solving this and how I would think about it.
You first put down the $1$'s in $5$ positions. There are $30\choose{5}$ ways to do this.
You next put down the $2$'s in $5$ positions. There are $25\choose{5}$ ways to do this.
You next put down the $3$'s in $5$ positions. There are $20\choose{5}$ ways to do this.
You next put down the $4$'s in $5$ positions. There are $15\choose{5}$ ways to do this.
You next put down the $5$'s in $5$ positions. There are $10\choose{5}$ ways to do this.
Finally, you put down the $6$'s in $5$ positions. There are $5\choose{5}$ ways to do this.
Multiplying all these together, you get $\approx 8.88\cdot10^{19}$
 which is the same thing you'd get from doing $$\frac{30!}{(5!)^6}$$
Then multiply by $\frac{1}{6}^{30}$, since there is a probability of $\frac{1}{6}$ that a given number gets put in a given slot, to get 
$${\frac{30!}{(5!)^6}\cdot\frac{1}{6}}^{30}\approx 4.02\cdot10^{-4}$$
A: Here's a simpler way of looking at this problem. There are $6^{30}$ possibilities. To count he favorable ones, treat the rolls as 30 characters that can be rearranged: 111112222233333...66666
There are $30!/(5!)^6$ arrangements which leads us to our answer $30!/(5!^6\times6^{30})$.
