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Anit [1.1K]
3 years ago
15

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?
Mathematics
1 answer:
mestny [16]3 years ago
6 0

Answer:

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

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Answer:

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Step-by-step explanation:

The ratio of Ed's toy car to Pete's toy car is initially given as 5:2

Ed gave Pete a total number of 30 cars

Let x represent the greatest common factor that exists between both number

Number of Ed's car is represented as 5x

Number of Pete car is represented as 2x

Since they each have an equal number of cars which is 30 then we can solve for x as follows

5x-30=2x+30

Collect the like terms

5x-2x= 30+30

3x= 60

Divide both sides by the coefficient of x which is 3

3x/3=60/3

x=20

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Therefore, the total number of cars can be calculated as follows

= 100+40

= 140 toy cars

Hence they have 140 toy cars altogether

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