# permutation 3

• Feb 6th 2009, 12:38 PM
william
permutation 3
1. How many 3-letter arrangements are there of the letters of the word CANADA?
• Feb 6th 2009, 12:49 PM
Plato
Quote:

Originally Posted by william
1. How many 3-letter arrangements are there of the letters of the word CANADA?

If the word were $\displaystyle C A_1 N A_2 D A_3$ how many ways would there be?
• Feb 6th 2009, 01:03 PM
william
Quote:

Originally Posted by Plato
If the word were $\displaystyle C A_1 N A_2 D A_3$ how many ways would there be?

If I use 1 A it will be 4!=24

2A's it'll be 9

and 3A's it'll be 1

so which of these are correct?
• Feb 6th 2009, 01:06 PM
Plato
NO!
You did not answer the question. With the subscripts there are six different letters. Now answer.
• Feb 6th 2009, 01:10 PM
william
Quote:

Originally Posted by Plato
NO!
You did not answer the question. With the subscripts there are six different letters. Now answer.

if we consider the numbers the A's as different letters it's 6!=720 there are 6 ways the A's can be arranged (3!) so 720/6=120?
• Feb 6th 2009, 01:12 PM
Plato
Quote:

Originally Posted by william
if we consider the numbers the A's as different letters it's 6!=720 there are 6 ways the A's can be arranged (3!) so 720/6=120?

GREAT By George he got it.

How many ways can you arrange "MISSISSIPPI"?
• Feb 6th 2009, 01:16 PM
william
Quote:

Originally Posted by Plato
GREAT By George he got it.

How many ways can you arrange "MISSISSIPPI"?

I'm guessing you take 11! ( number of letters in the word ) over 4!( 4s's) 4!(4i's) and 2!(2p's)

so $\displaystyle \frac{11!}{4!4!2!}$ is what I'm trying to say, but I am having trouble computing this in my calculator
• Feb 6th 2009, 01:20 PM
Plato
Quote:

Originally Posted by william
I'm guessing you take 11! ( number of letters in the word ) over 4!( 4s's) 4!(4i's) and 2!(2p's)
so $\displaystyle \frac{11!}{4!4!2!}$ is what I'm trying to say, but I am having trouble computing this in my calculator

CORRECT!
See how much one can learn if answers are not just passed out.
• Feb 6th 2009, 01:24 PM
william
Quote:

Originally Posted by Plato
CORRECT!
See how much one can learn if answers are not just passed out.

Absolutely, I prefer you giving me hints rather then giving me the answers, and I appreciate it. Thanks for your time!(Hi)
• Feb 8th 2009, 08:53 AM
william
Quote:

Originally Posted by Plato
GREAT By George he got it.

How many ways can you arrange "MISSISSIPPI"?

I just checked the answer in the back of the book and it says 34?
• Feb 10th 2009, 11:55 AM
Nisar_0926
permutation
Most of the answers at the back of the textbook are wrong.
• Feb 10th 2009, 12:02 PM
TheMasterMind
Quote:

Originally Posted by william
If I use 1 A it will be 4!=24

2A's it'll be 9

and 3A's it'll be 1

so which of these are correct?

This is actually correct! and the answer in your book is correct!

Splitting it up into three different cases

for 1A is is 4!=24

2A's 3*3=9
(Ex.DAA)

3A's=1
(Ex. AAA)

Thus using the additive principle associated with permutations add all three cases together: $\displaystyle \boxed{24+9+1=34}$