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

you ignore the remainder, therefore the answer is y = x+2

step-by-step explanation:

you have to divide the denominator by the numerator. you can use synthetic division or long division.

[-5] .     1 .     7 .     11

x + 2 + 1/x+5