Eight clerks and 5 drivers apply for positions in a company.?
This is the same as Sergie Michallovs answer, but with more detail. The question involves both union and intersection of events, but this is not a helpful way to compute the answer. The better way to solve this problem is to count all the possible sets of four applicants, and also how many of them consist of two clerks and two drivers. To select four applicants, we can pick the first one in 13 ways, the second one in 12, the third in 11 and the fourth in 10. Thats 13\times12\times11\times10=17,160 ways. But not all those selections are different. For example we might select applicant 1, followed by 2, 3 and 4but thats the same as selecting applicant 2, followed by 1, 3 and 4, or first selecting 3 followed by 4, 2 and 1. there are 4\times3\times2\times1=24 ways to reorder four accepted applicants. So there are \frac{17,160}{24}=715 different sets of four applicants you could select out of 13. This is called 13 choose 4 and is written \binom{13}{4}. We use the same method to figure out how many ways we can select two clerks out of 8, \binom{8}{2}=\frac{8\times7}{2\times1}=28, and how many ways to select two drivers out of 5, \binom{5}{2}=\frac{5\times4}{2\times1}=10. Multiply them to get 28\times10=280 ways to select two clerks and two drivers. Imagine instead of picking applicants one at a time, all 715 possible selections of four applicants are written on slips of paper, and put into a hat. 280 slips contain two clerks and two drivers. If a slip is drawn at random, the chance is \frac{280}{715} that it has two clerks and two drivers.