Expert: Sherry Wallin - 8/18/2010

Question
a student has to answer 10 questions, choosing at least 4 from
each of part A and part B. If there are 6 questions in part A
and 7 in part B, in how many ways can the student choose 10
questions ?

Answer
Nithin~

The only sums of 10 that fit the criteria are 4 + 6, 5 + 5, 6 + 4
You can think of the sums as being a + b so the first one is 4 from A and 6 is from B. The 3rd sum is 6 from A and 4 is from B. I make this point so you can see all the cases are accounted for.

I will use the notation nCr to mean 'n things taken r at a time'.
I am also going to assume you know how to use factorials with combinations, if this is not the case please write me back. I will work the first sum out for you with explanations and I will also tell you (if you don't already know this) that 0! = 1! = 1

6C4 * 7C6
= 6!/(2!4!) * 7!/(1!6!)
= (6*5*4!/(2!4!) * (7*6!)/(1!6!)  
= (6*5/1*2)* 7/1          (the 4 factorials and 6 factorials canceled)
= 30/2 * 7
= 15*7
= 105

This means there are 105 ways to choose 4 questions from part A and choose 6 questions from part B.

from sum #2:  6C5 * 7C5 = 6*21 = 126 ways to choose 5 questions from part A and choose 5 questions from part B

from sum #3:  6C6 * 7C4 = 1 * 35 = 35 ways to choose 6 questions from part A and choose 4 questions from part B.


So the total ways to choose 10 questions from a pool of 6 questions from a and 7 questions from B where you have to have at least 4 questions from each is 105 + 126 + 35 = 266 total ways

The multiplication principle is why we are multiplying each of the combinations, for example in 6C4 * 7C6, rather than adding them we know that for each of the 15 ways of choosing 4 questions from part A that there are 7 ways to choose 6 questions from part B.

Math Prof