Combinatorics - Choose Any Marble Exactly 4 times Suppose a bag contains 10 marbles all uniquely distinct from one another and we are to pick 1 marble out of the bag a time 10 times with replacement. 
What are the odds that you pick the same marble 4 times? 3 times? 2 times? Every marble exactly once?

I'm going to focus on the P(4) case which is the probability that we pick any marble exactly 4 times. 
I think what is tripping me up is the end part of eliminating special cases. 
I know for reference that unordered sampling with replacement there are ${n+k-1\choose k} = {10+10-1\choose 10} = {19\choose 10}\ = 92, 378$ total combinations.
How I was trying to solve it was take the probability of getting the same marble 4 times is $\frac{1}{10} ^4 * 10$ (since there are 10 marbles that could be repeated 4 times. There are then $_{10}C_4$ combinations for getting 4 of the same marble so ${10\choose 4} = 210 $ total combinations with 4 the same. 
Then I was just going to multiply by  $\frac{9}{10} ^6$ for .001*210*.531441 = 11.16 but I realize that out of the 6 remaining marbles I can't pick the same one more than 4 times for this to be truly accurate. And this method doesn't seem to apply correctly for P(3) and P(2) so I must be doing something wrong. 
Thanks for the help!
 A: Order does matter here. If you want every element of your probability space to be equally likely, you need to count ordered sequences, not unordered ones. For example, the probability of observing the same marble 10 times is not the same as the chance of observing a particular marble nine times and a different one once. The former has probability $\frac{1}{10^{10}}$, the latter $\binom{10}1 \cdot \frac1{10^{10}}$.

I think you need to do inclusion-exclusion to answer this. Number the marbles 1 to 10, and for each $i$, find the probability of choosing marble $i$ four times. There are $\binom{10}4$ ways to choose which times you draw the $i$, then $9^6$ ways to choose the other marbles. Since there are $10$ choices for $i$, it seems like the probabability would be 
$$
10\cdot \frac{\binom{10}49^6}{10^{10}}
$$
However, sequence which have two marbles $i$ and $j$ both appearing four times are double counted by the above expression. We must subtract these out, resulting in 
$$
10\cdot \frac{\binom{10}49^6}{10^{10}}-\binom{10}{2}\frac{\binom{10}4\binom{6}48^2}{10^{10}}
$$
