# Permutation

How to solve a permutation (_{n}P_{r}): formula, 4 examples, and their solutions.

## Formula_{n}P_{r}

Permutation _{n}P_{r} means

start from n and multiply r numbers.

(n ≥ r)

Special values:_{n}P_{1} = n_{n}P_{0} = 1

## Example_{6}P_{4}

_{6}P_{4} is,

start from 6 and multiply 4 numbers,

6⋅5⋅4⋅3.

6⋅5 = 30

4⋅3 = 12

30⋅12 = 360

So 360 is the answer.

## ExampleWord Problem

_{n}P_{r} is used

when choosing r things from n things

and arranging them in a row.

It says

choose 3 students from 9 students

and arrange them in a row.

So the number of the ways is_{9}P_{3}.

_{9}P_{3} is,

start from 9 and multiply 3 numbers,

9⋅8⋅7.

9⋅8 = 72

72⋅7 = 504

So 504 is the answer.

## ExampleWord Problem

There are 8 numbers.

To make a 4-digit number,

choose 4 numbers

and arrange them in a row.

So the number of the ways is_{8}P_{4}.

_{8}P_{4} is,

start from 8 and multiply 4 numbers,

8⋅7⋅6⋅5.

8⋅7 = 56

6⋅5 = 30

56⋅30 = 1680

So 1680 is the answer.

## ExampleWord Problem

There are 6 numbers.

To make a 3-digit number,

choose 3 numbers

and arrange them in a row.

The number of the ways is_{6}P_{3}.

But there's a case to subtract.

If 0 is in the hundreds,

then it's not a 3-digit number:

[012] is not a 3-digit number.

So you should subtract this case.

If 0 is in the hundreads,

there are 5 numbers to choose,

choose 2 numbers,

and arrange them in tens and ones.

The number of the ways to subtract is_{5}P_{2}.

So the number of ways

to make a 3-digit number is_{6}P_{3} - _{5}P_{2}.

_{6}P_{3} is,

start from 6 and multiply 3 numbers,

6⋅5⋅4.

Minus,_{5}P_{2} is,

start from 5 and multiply 2 numbers,

5⋅4.

So _{6}P_{3} - _{5}P_{2}

= 6⋅5⋅4 - 5⋅4.

6⋅5⋅4 - 5⋅4

= 5⋅4(6 - 1)

Common Monomial Factor

5⋅4 = 20

6 - 1 = 5

20⋅5 = 100

So 100 is the answer.