There are 10 computers and 5 students. How many ways are there for the students to sit at the computers if no computer has more than one student and each student is seated at a computer?

30,240 ways

Step-by-step explanation:

This question is bothered on permutation. Permutation has to do with arrangement.

If there are 10 computers and 5 students, the number of ways students will sit at the computers if no computer has more than one student can be expressed as;

10P5 = 10!/(10-5)!

10P5 = 10!/(5)!

10P5 = 10*9*8*7*6*5!/5!

10P5 = 10*9*8*7*6

10P5 = 30,240

Hence the number of ways is 30,240 ways

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

y > 1/3x

y < = -x + 4

hope this : )