There are 20 identical printing jobs to be sent to 5 different printers. (a) one of the printers is much faster than the other four. how many ways are there to distribute the jobs if the faster printer gets at least 7 jobs? (b) one of the printers is almost out of paper. how many ways are there to distribute the jobs so that the printer that is almost out of paper does not get more than three jobs? (in addition to the fact that the faster one gets at least 7 jobs)

a) 482,718,652,416,000

b) 80,453,108,736,000

Step-by-step explanation:

(a)

There are (combinations of 20 taken 7 at a time)

different ways of sending 7 jobs to the fastest printer.

There are

13! = 6,227,020,800

ways of sending the other 13 jobs to the rest of the printers (assuming that none of the printers is idle)

By the Fundamental Rule of counting, there are  

77520*6,227,020,800 = 482,718,652,416,000

different ways of distributing the jobs with the faster printer getting at least 7 jobs.

(b)

There are (combinations of 20 taken 7 at a time)

different ways of sending 7 jobs to the fastest printer.

There are (combinations of 13 taken 3 at a time)

different ways of sending 3 jobs to the printer that is almost out of paper.

There are  

10! = 3,628,800

different ways of sending the 10 remaining jobs to the rest of the printers.

By the Fundamental Rule of counting, there are  

77,520*286*3,628,800 = 80,453,108,736,000

different ways to distribute the jobs so that the printer that is almost out of paper does not get more than three jobs and the faster one gets at least 7 jobs.

the answer is c: divide the arc's degree measure by 360°, then multiply by the circumference of the circle.