another question concerning the binomial coefficient from my "discrete mathematics and its applications" book:
"there are 8 sorts of distinguishable devices. How many ways are there to choose a dozen devices with at least 3 of sort A and no more than 2 of sort B." solution: 9724
a dozen = 12. 12 - 3 ( at least 3 of sort A ) = 9. 9-2 = 7 - so i can choose 7 out of 7 sorts, this is the first part of the sum. the second part of the sum is my second choise of 2 out of 8 distinguishable devices. i sum them up because there is no intersection.
what did i do wrong ?