Question 887210
the 10 tv sets include 7 good and 3 defective
to get at least 2 defective in a set of 4, the dealer needs to get sets of 4 tv's that include exactly 2 defectives each and sets of 4 tv's that include exactly 3 defectives each.


each set of 2 defectives will include 2 good and 2 defectives.
each set of 3 defectives will include 1 good and 3 defectives.


the number of ways you can get a set of 2 defectives and 2 good is equal to:
3C2 * 7C2


the number of ways you can get a set of 3 defectives and 1 good is equal to:
3C3 * 7C1


3C2 * 7C2 = 3 * 21 = 63


3C3 * 7C1 = 1 * 7 = 7


total number of ways to get sets of 4 that include at least 2 defectives each is equal to 63 + 7 = 70.


nCx formula is equal to n! / (x! * (n-x)!


for example:


7C2 = 7! / (2! * 5!) = (7*6*5*4*3*2*1) / (2*1*5*4*3*2*1) = 21