Question

how many ways are there of choosing
2
letters, without replacement, from the
5
letters A, B, C, D, and E, if
the order of the choices matters?
the order of the choices does not matter?

Answer 1

In these types of problems, if the order matters, it's a permutation problem.
If the order does not matter, it's a combination problem.

Answer 2

well it si asking for both..... I dont understand how to do it and my math teacher did not teach this at all in class..... so

Answer 3

In the permutation problem, you need to choose all permutations of 2 letters from 5.
That is \(\large_5P_2\).
The formula for permutation is:
\(\large _nP_r = \dfrac{n!}{(n - r)!} \)

Answer 4

WHAT :#

Answer 5

You are choosing all permutations of 2 items chosen from 5.
In your case, n = 5, and r = 2.
Just put them in the formula above and evaluate it.
Are you familiar with factorial numbers?

Answer 6

NNOOOO

Answer 7

!!!!!

Answer 8

Ok. Then let me explain that to you.
A factorial number is this:
1! (read as one factorial), 2! (read as 2 factorial) etc.
The definition is 1! = 1
2! = 2 * 1 = 2
3! = 3 * 2 * 1 = 6
4! = 4 * 3 * 2 * 1 = 24
etc.

Answer 9

oh... so
7!: 1,7
and
6!: 1,2,3,6

Answer 10

An integer greater than 1 factorial is equal to that integer multiplied by all integers less than that number and greater than or equal to one.

Answer 11

or is it.....

Answer 12

No. Let's do one thing at a time.
For now we are just learning about factorial.
Then we will go over the formula for permutations.

Answer 13

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040

Answer 14

so....................
7!: 7*6*5*4*3*2*1
and
5!: 5*4*3*2*1

Answer 15

6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

Answer 16

Correct. Now you got it!

Answer 17

YAY

Answer 18

so

Answer 19

Ok. That is how factorial numbers work.
Now we can go on to the formula for the number of permutations.
If you are choosing 2 items out of 5, where order matters, the formula is:
\(\large _nP_r = \dfrac{n!}{(n - r)!}\)
for choosing r items out of n items where order matters.

Answer 20

In your case, n = 5, and r = 2, so we can plug in those numbers in the formula to get:
\(\large _5P_2 = \dfrac{5!}{(5 - 2)!}\)
Now we need to simplify the formula until we get a number.

Answer 21

\(\large _5P_2 = \dfrac{5!}{(5 - 2)!} = \dfrac{5!}{3!} = \dfrac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}\)

Answer 22

Ok so far?

Answer 23

yeah.

Answer 24

We need a number at the end, not an operation, so we need to continue simplifying.

Answer 25

so it's 120/6

Answer 26

20

Answer 27

\(=\dfrac{5 \times 4 \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\cancel{3} \times \cancel{2} \times \cancel{1}} = 5 \times 4 = 20\)

Answer 28

Correct.

Answer 29

Now we need to do the second part.
If order does not matter, then we will get fewer results.
Then we are dealing with a combination problem.
The formula for the number of combibnations of choosing r items from n items is:

\(_nC_r = \dfrac{n!}{(n - r)!r!} \)

Answer 30

Now try this formula with n = 5 and r = 2, and see what you get.

Answer 31

@mathstudent55 I am sorry I had to eat dinner ;) lol

Answer 32

No problem. Just try the combination formula with n = 5 and r = 2, and let me know what you get.