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olga nikolaevna [1]
2 years ago
12

If 5 items are chosen at a random without replacement from a list of 10 items, in how many ways can the 5 items be arranged if t

he order of items in arrangement is important?
Mathematics
1 answer:
Mashcka [7]2 years ago
5 0

Considering the definition of permutation, the items can be arranged in 30,240 ways.

<h3>What is Permutation</h3>

Permutation is placing elements in different positions. So, permutations of m elements in n positions are called the different ways in which the m elements can be arranged occupying only the n positions.

That is, permutations refer to the action of arranging all the members of a set in some sort of order or sequence.

In other words, permutations (or Permutations without repetition) are ways of grouping elements of a set in which:

  • take all the elements of a set.
  • the elements of the set are not repeated.
  • order matters.

To obtain the total of ways in which m elements can be placed in n positions, the following expression is used:

mPn= m!÷ (m-n)!

where "!" indicates the factorial of a positive integer, which is defined as the product of all natural numbers before or equal to it.

<h3>Ways the items can be arranged</h3>

In this case 5 items are chosen at a random without replacement from a list of 10 items.

You want to know in how many ways can the 5 items be arranged if the order of items in arrangement is important. So, you use the permutation 10P5.

10P5= 10!÷ (10-5)!

Solving:

10P5= 3,628,800÷ 5!

10P5= 3,628,800÷ 120

<u><em>10P5= 30,240</em></u>

Finally, the items can be arranged in 30,240 ways.

Learn more about permutation:

brainly.com/question/12468032

brainly.com/question/4199259

#SPJ1

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Step-by-step explanation:

(see attached for reference)

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