Question

MEDAL!

How many ways can you arrange the word REFEREE?

Answer 1

to make it spell referee again?

Answer 2

I don't think so. We need to use permutations to solve this problem I think.

Answer 3

@agent0smith

Answer 4

@robtobey

Answer 5

I thought that we could use:
\[7\text{P}7 = \frac{ 7! }{ (7 - 7)! } = 5,040\]
But the answer key says 105. I don't know why.

Answer 6

You also have to divide by the factorial of all the repeated letters. There's two R's so 2!, and four E's so 4!

Answer 7

I don't understand.

Answer 8

I mean why do I need to divide by the factorial of all the repeated letters?

Answer 9

Because what difference is there between EEEE and EEEE, or EEEE? There's four E's, so 4! ways to arrange those four E's... but they're all the same.

Answer 10

So you're saying that whenever I have a problem like this, I should divide the factorial of the number of times a letter is repeated with the total permutation?

Answer 11

Like R is repeated 2 times so it would be 2! and E is repeated 4 times so it would be 4!. The total permutation is 5,040, but I would now need to do \(\large \frac{5,040}{2! \times 4!}\).

Answer 12

Would that be correct?

Answer 13

Yep!

Answer 14

Okay. I am going to solve the problem now.

Answer 15

Thank you so much!

I also just wanted to make sure that I can easily distinguish a permutation from a combination. How can I do that?

Answer 16

It takes practice.

But permutation means that the order of arrangement matters. Like if you're choosing 3 people to win a prize, that's a combination since they all get the same prize.

If you're choosing 3 people to win 1st, 2nd, 3rd prizes which are different, then the order matters.

Answer 17

no. of R's=2
no. of E's=3
total letters=7
no. of words
\[=\frac{ 7! }{ 2!*3! }\]

Answer 18

I have another question relating to this topic:

How many permutations can occur by rearranging the letters in the word, CHAPTER if:
a) The last letter must be a vowel?

Answer 19

correction
no. of E's=4
no. of words
\[=\frac{ 7! }{ 2!*4! }\]

Answer 20

There's only two vowels, so just make two possibilities (there's no repeated letters to worry about)

_ _ _ _ _ _ E

_ _ _ _ _ _ A

The blank spaces are where letters can go. We've already used 1 letter, so there's 6 letters left. Then the next spot has 5, then 4 and so on. Multiply.

Answer 21

Since there are two vowels can we have written:
2 x 6 x 5 x 4 x 3 x 2 x 1 ?

Answer 22

Yes.

Answer 23

I have some other questions relating to permutations, can you help me?

Answer 24

I gotta leave.

Answer 25

Okay. Thank you for your help.

Answer 26

You're welcome.