i dont know exactly how to solve this problem:
a classroom has two rows of 8 seats each. there are 14 students, 5 of whom always sit in the front row and 4 of whom sit in the back row. in how many ways can the students be seated? help?
a classroom has two rows of 8 seats each. there are 14 students, 5 of whom always sit in the front row and 4 of whom sit in the back row. in how many ways can the students be seated? help?
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since the students have names, order matters, so permutations instead of combinations.
8p5 = 8•7•6 = 336 ways to seat 5 in front.
8p4 = 8•7•6•5 = 1680 ways to seat 4 in back.
that uses 9 of the 16 seats, so
7p5 = 7•6 ways to seat the remaining 5 students, for a total of
336 • 1680 • 42 = 23,708,160 ways to seat the students.
8p5 = 8•7•6 = 336 ways to seat 5 in front.
8p4 = 8•7•6•5 = 1680 ways to seat 4 in back.
that uses 9 of the 16 seats, so
7p5 = 7•6 ways to seat the remaining 5 students, for a total of
336 • 1680 • 42 = 23,708,160 ways to seat the students.