Probability question regarding coin flips. There are 20 coins in a jar. Of these, 8 are quarters, 5 are dimes, 3 are nickels, and 4 are pennies. 8 coins are drawn at random, without replacement from the jar.
What is the chance that the fourth coin is a quarter and the eighth coin is a dime?
For this question I used combination method and I got: 
$\frac{\binom{8}{4}\binom{5}{1}}{\binom{20}{8}}$
which is equal to 0.0028
The correct answer is 0.105
I don't know what I did wrong here.
What is the chance that the last two coins are of the same denomination?
for this question I got:
P(last 2 coins are quarters)+P(last 2 coins are dimes)+P(last two coins are nickels)+P(last two coins are pennies)
$\frac{\binom{12}{6}\binom{8}{2}+\binom{15}{6}\binom{5}{2}+\binom{17}{6}\binom{3}{2}+\binom{16}{6}\binom{4}{2}}{\binom{20}{8}}$
is this correct?
Thank you!

What is the chance that the fourth coin is a quarter and the eighth coin is a dime?
There is actually a simpler way but let’s push forward with your combinatorics method. For the first 4, the cases are that there may be 1, 2, 3, or 4 quarters by the time the fourth coin is picked so we need to take care of that. We don't have to care about quarters from the 5th draw onwards.
Similarly, amongst the first 8 (except 4th) we also have to take note that there may be anything from 1 to 7 dimes in picked by the eighth draw.
Now let's look at some cases
Case A | No quarter, no dime excluding on 4th and 8th respectively.
$$p_A=\frac{8}{17}\cdot\frac{5}{13}\cdot\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2}{20\cdot19\cdot18\cdot16\cdot15\cdot14}$$
Case B | 1 quarter, no dime excluding on 4th and 8th respectively.
$$p_B=\frac{8-1}{17}\cdot\frac{5}{13}\cdot\frac{\binom31(8)(6\cdot5\cdot4\cdot3\cdot2)}{20\cdot19\cdot18\cdot16\cdot15\cdot14}$$
Case C |  2 quarters, 3 dimes excluding on 4th and 8th respectively.
$$p_C=\frac{8-2}{17}\cdot\frac{5-3}{13}\cdot\frac{\binom32(8\cdot7)\binom{6-2}3(5\cdot4\cdot3)(7)}{20\cdot19\cdot18\cdot16\cdot15\cdot14}$$
We can then  generalize this to the case with $n$ quarters and $m$ dimes (outside of 4th and 8th places) whose probability is
$$p_{n,m}=\frac{8-n}{17}\cdot\frac{5-m}{13}\cdot\frac{\binom3n\frac{8!}{(8-n)!}\binom{6-n}{m}\frac{5!}{(5-m)!}\frac{7!}{(7-(n+m))!}}{20\cdot19\cdot18\cdot16\cdot15\cdot14}$$
From here onwards, it's tedious work, or you can use the summation:
$$\text{Probability required}=\sum_{n=0}^{3}{\sum_{m=0}^{6-n}{p_{n,m}}}$$ to help you out!

What is the chance that the last two coins are of the same denomination?
You were almost correct but your approach would require you to take care of the permutation (i.e. different orders the coins can be) and the identical nature of the coins with the same denominations. An easier approach would be to do independent choosing for each placing! I'll let you try out yourself first and will put up the full approach in a while.
