You mistyped since 10 is not 6 + 9.  The problem was either

1. There is a true-false quiz with 10 questions. How many possible answer keys
are there if the answer to 6 questions are false and the other 4 questions are
true?

or

2. There is a true-false quiz with 15 questions. How many possible answer keys
are there if the answer to 6 questions are false and the other 9 questions are
true?

I'll assume you meant 2:

Here is a typical answer key:

1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
T   F   F   T   T   T   F   T   F   T   T   F   T   F   T

We can work it either of two ways:

I. By placing all the T's first:

There are "15 choose 6" ways to choose the numbers to put T's.  There is
only one way to fill in all the others, since they are all F's  So the answer
is 15C6 = 5005 possible answer keys.

 or

II. By placing all the F's first:

There are "15 choose 9" ways to choose the numbers to put F's.  There is
only one way to fill in all the others, since they are all T's  So the answer
is 15C9 = 5005 possible answer keys.

The answers are the same because 15C6 and 15C9 are the same. That's because the
number of ways to choose 6 out of 15 is the same as the way to leave 9
behind.

-----------------------------------
-----------------------------------

We have 10 question numbers like this:

1   2   3   4   5   6   7   8   9  10  


There are "10 choose 2" or 10C2 question numbers to place the A's.
Here is one of them:

1   2   3   4   5   6   7   8   9  10  
        A               A         
For each of the 10C2 ways to place the A's, in each case there will
remain only 8 places to put the others. So, having placed the A's,
there are "8 choose 3" or 8C3 question numbers to place the B's under.
That's 10C2*8C3 ways to place the A's and B's.  Here is one of them:

1   2   3   4   5   6   7   8   9  10  
B       A   B           A       B   

For each of the 10C2*8C2 ways to place the A's and B's, in each case 
there will remain only 5 places to put the others. So, having placed 
the A's and B's, there are "5 choose 2" or 5C2 question numbers to place
the C's under.  That's 10C2*8C3*5C2 ways to place the A's, B's, and C's.
Here is one of them:

1   2   3   4   5   6   7   8   9  10  
B       A   B       C   A       B   C

For each of the 10C2*8C2*5C2 ways to place the A's, B's, and C's, in each
case there will remain only 3 places to put the D's. So, having placed
the A's, B's, and C's there are "3 choose 3" or 3C3 question numbers to place
the D's under. [NOTICE THAT 3C3 IS JUST 1 because there is only one way to
put D's under all 3].  That's 10C2*8C3*5C2*3C3 ways to place the A's, B's, C's,
and D's.  That's 25200 ways.  Here is one of them completed
 
1   2   3   4   5   6   7   8   9  10  
B   D   A   B   D   C   A   D   B   C

Notice that it would not have mattered which letter we placed first.
The numbers to multiply may be different, but the answer would always 
be the same.  Here are other possibilities of placing 

10C2*8C3*5C2*3C3 =  45*56*10*1 = 25200, placing the A's, then B's, then C's, then D's

10C2*8C3*5C3*2C2 =  45*56*10*1 = 25200, placing the A's, then B's, then D's, then C's
 
10C2*8C2*6C3*3C3 =  45*28*20*1 = 25200, placing the A's, then C's, then B's, then D's
 
10C3*7C2*5C3*2C2 = 120*21*10*1 = 25200, placing the B's, then A's, then D's, then C's
 
10C3*7C2*5C2*3C3 = 120*21*10*1 = 25200, placing the B's, then A's, then C's, then D's
 
10C3*7C3*4C2*2C2 = 120*35*6*1  = 25200, placing the B's, then D's, then A's, then C's
 
Edwin