Combination of a combination I have an out of curiosity question that's been stumping me for 2 days. If there is a quiz comprised of 10 questions randomly chosen from a 15 question bank, how many quizzes will have at least 7 questions in common? I know $\binom{15}{10}$ yields the number of possible unique quizzes (3003), but I can't get it to the next stage of figuring out how many of those quizzes would have 7 questions in common (or 8 or 9 or 10) to figure out the total probability. Could someone possibly give me a hint as to how to think about this?
 A: One way to approach this is to create quiz 1 and a quiz 2 and see in how many ways you can do that so that they have $7$ or more questions in common. And to do that, split it up into cases: exactly $7$ in common, exactly $8$ in common etc.
Now, for quiz 1 and quiz 2 to have exactly 7 questions in common, for quiz 1 to have $3$ more questions in addition to those $7$, and for quiz 2 to also have $3$ more in addition,  but different from the $3$ that quiz 1 has: First pick 7 out of 15 that the havein common, then pick 3 out of the remaining 8 for quiz 1, and finally 3 out of the remaining 5 for quiz 2. So, the number of ways to create a quiz 1 and a quiz 2 with exactly $7$ questions in common is:
$${{15}\choose{7}}* {8\choose3}*{5\choose3}$$
Similarly, the number of ways to create quiz 1 and quiz 2 that have exactly 8 questions in common is:
$${{15}\choose{8}}* {7\choose2}*{5\choose2}$$
Exactly 9 in common:
$${{15}\choose{9}}* {6\choose1}*{5\choose1}$$
Exactly 10 in common:
$${{15}\choose{10}}* {5\choose0}*{5\choose0}$$ 
Add them up and you have the number of ways you can create quiz 1 and quiz 2 with 7 or more questions in common.  
This is out of 
$${{{15}\choose{10}}}*{{15}\choose{10}}$$
possible ways to create quiz 1 and quiz 2, so divide by that, and you have the probability of getting two quizzes with $7$ or more questions in common.
