I have 10 red and 10 blue cards. I shuffle the cards and then label the cards based on their orders: I write the number one on the first card, the number two on the second card, and so on. What is the probability that
a) All red cards are assigned numbers less than or equal to 15? b) Exactly 8 red cards are assigned numbers less than or equal to 15?
I know that the total number of ways to arrange 20 cards is $20!$. How do I proceed further?
Let the number of red cards be $N_1$ and that of blue ones $N_2$ (in your case $N_1=N_2=10$) and ask in how many permutations $k$ red cards will be among the $K$ first.
The answer is $$ \binom {N_1}k\binom {N_2}{K-k}K!(N_1+N_2-K)! $$ where $\binom {N_1}k\binom {N_2}{K-k}$ stays for the number of ways to choose $k$ red cards and $K-k$ blue ones, $K!$ - for the number of ways to permute them, and $(N_1+N_2-K)!$ - for the number of ways to permute the rest cards. It remains to divide the result by $(N_1+N_2)!$ to obtain: $$ p=\frac{\binom{N_1}{k}\binom{N_2}{K-k}}{\binom{N_1+N_2}K}. $$ This result can be of course obtained also directly by counting the respective combinations.