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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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DaniilM [7]
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5 0
3 years ago
Divide 7/24 by 35/48 and reduce the quotient to the lowest fraction
tiny-mole [99]
\frac{7}{24} ÷ \frac{35}{48}
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We can simplify it further:
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4 0
3 years ago
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8 0
3 years ago
Read 2 more answers
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kompoz [17]
So, you have 3/4x + 5/8 = 4x. We can do this in two ways. Since fractions are harder to deal with, we can turn the fractions into whole numbers.

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