# Thread: [SOLVED] Permutations Problem

1. ## [SOLVED] Permutations Problem

The language with the most letters in it is Cambodian, which has 72 letters.

b) How many three-letter permutations can be created if the letters may be repeated?

The options would look something like this: AAABBBCCCDDDEEEFFF...

So I tried: $\displaystyle \frac{72!}{72(3!)}$ but that was too big for my calculator.

I don't know how else to solve this.

Any help is greatly appreciated.

2. Originally Posted by Some_One
The language with the most letters in it is Cambodian, which has 72 letters.

b) How many three-letter permutations can be created if the letters may be repeated?

The options would look something like this: AAABBBCCCDDDEEEFFF...

So I tried: $\displaystyle \frac{72!}{72(3!)}$ but that was too big for my calculator.

I don't know how else to solve this.

Any help is greatly appreciated.
$\displaystyle ^nP_r = \frac{n!}{(n-r)!}$

n = 72 and r = 69

$\displaystyle ^nP_r = \frac{72!}{(72-3)!} = (72) \times (71) \times (70)$

3. Hello, Some_One!

The language with the most letters in it is Cambodian, which has 72 letters.

b) How many three-letter permutations can be created if the letters may be repeated?

The options would look something like this: AAABBBCCCDDDEEEFFF... . what?

So I tried: $\displaystyle \frac{72!}{72(3!)}$ .??
These are three-letter "words" like: .$\displaystyle AAA, AAB, AAC\; \hdots\; ZZZ$

There are 72 choices for the first letter.
. . There are 72 choices for the second letter.
. . . . There are 72 choices for the third letter.

Therefore . . .

4. Originally Posted by e^(i*pi)
$\displaystyle ^nP_r = \frac{n!}{(n-r)!}$

n = 72 and r = 69

$\displaystyle ^nP_r = \frac{72!}{(72-3)!} = (72) \times (71) \times (70)$
That would be if the letters may NOT be repeated.

My problem is if letter CAN be repeated.

5. Originally Posted by Soroban
Hello, Some_One!

These are three-letter "words" like: .$\displaystyle AAA, AAB, AAC\; \hdots\; ZZZ$

There are 72 choices for the first letter.
. . There are 72 choices for the second letter.
. . . . There are 72 choices for the third letter.

Therefore . . .

Doh! Of course! I can't believe that I forgot that.

Before this question I was doing a problem that involved finding the permutations of A,B,D,D,D so I still had that mind set.

Thanks for your help!

6. Hello again, Some_One!

Before this question I was doing a problem that involved finding
the permutations of A,B,D,D,D so I still had that mind set.
Boy, do I know that feeling!

I have this flat spot on my forehead to prove it.