# Permutation

• Oct 12th 2008, 02:53 AM
Geometor
Permutation
Hi, just to make sure I got the right idea.

In how many ways can the letters of the word MERETRICIOUS be arranged in a line?

Since there are 2E, 2I and 2R. Would the first answer be 12! / (2! 2! 2!)?

What is the probability that an arrangement begins with M and ends with S.

Would the probability be (10!/(2! 2! 2!) / (12!/ (2! 2! 2!) as a result?

Thanks for any clarification.
• Oct 12th 2008, 08:17 AM
jkru
The first part is right.

I don't think the second part is right though. The probability of choosing an M as the first letter is going to be 1choose(12) = n!/(n-1)!k!
So,

1choose(12) = 1!/11!12!

Now we want to choose an S as the last letter. The probability is now 1choose11 since we have already chosen an M and there are 1 fewer letters to choose. So,

1choose(11) = 1!/10!11!

Since we want both of these events to happen we multiply them together. To get the probability of this happening we divide by the total number of possibilities which you have calculated in the first part. So,

(1choose(12) * 1choose(11)) / (12!/(2!2!2!))

Hopefully this makes sense.
• Oct 12th 2008, 01:46 PM
Geometor
Choose? (Thinking)
I'm abit confused since I believe choose refers to combinations rather than permutations. And I think permutation are needed in this case since the order of arrangement does count as well

ah well, thanks for helping anyway :)
• Oct 12th 2008, 02:28 PM
Plato
Quote:

Originally Posted by Geometor
I'm abit confused since I believe choose refers to combinations rather than permutations. And I think permutation are needed in this case since the order of arrangement does count as well

Actually you first answer is correct.
$\displaystyle \frac{{\frac{{10!}} {{\left( 2 \right)^3 }}}} {{\frac{{12!}} {{\left( 2 \right)^3 }}}} = \frac{1} {{12\left( {11} \right)}}$
• Oct 13th 2008, 03:27 PM
bjhopper
Permutatons
originally posted by geometer

I agree with your solution with one exception.Shoudn't the numerator in the answer equation be 10 factorial divided by 16 to account for the fact that m or s can head the string requiring halving again

bj
• Oct 13th 2008, 05:17 PM
mr fantastic
Quote:

Originally Posted by Geometor
Hi, just to make sure I got the right idea.

In how many ways can the letters of the word MERETRICIOUS be arranged in a line?

Since there are 2E, 2I and 2R. Would the first answer be 12! / (2! 2! 2!)?

What is the probability that an arrangement begins with M and ends with S.

Would the probability be (10!/(2! 2! 2!) / (12!/ (2! 2! 2!) as a result?

Thanks for any clarification.

All your calculations are correct.