# Thread: Basic probability question with permutations

1. ## Basic probability question with permutations

What is the probability of these events when we randomly select a permutation of 1,2,....,n where n>=4

a) 1 precedes 2
I have no idea how to do this one

b) 1 immediately precedes 2
This one is simple I think. I just considered 1 and 2 as one unit to be permuted with the rest of the numbers and I get (n-1)!/n!= 1/n

I could really use some advice, thank you so much!!

2. Originally Posted by mulaosmanovicben
What is the probability of these events when we randomly select a permutation of 1,2,....,n where n>=4
a) 1 precedes 2
I have no idea how to do this one

b) 1 immediately precedes 2
This one is simple I think. I just considered 1 and 2 as one unit to be permuted with the rest of the numbers and I get (n-1)!/n!= 1/n

Part (a) is so easy!
Either 1 comes before 2 or not.

3. I am guessing answer to part a then is 1/2 for n>=4

4. Originally Posted by mulaosmanovicben
I am guessing answer to part a then is 1/2 for n>=4
Say n=2: 12; 21
Say n=3: 123; 132; 213; 231; 312; 321.

Does it matter if $\displaystyle n\ge 4?$

5. no it does not matter what n is. thank you i understand now. I am really having some difficulty with some of these problems.

n precedes 1 and n precedes 2

should this be 1/4 then? (either n precedes 1 or it doesnt and either n precedes 2 or it doesnt)

6. Originally Posted by mulaosmanovicben
n precedes 1 and n precedes 2
should this be 1/4 then? (either n precedes 1 or it doesnt and either n precedes 2 or it doesnt)
No it is $\displaystyle \frac{1}{3}$.

Look at 12n; 1n2; 21n; 2n1; n12; n21.
That is two out of six.

7. ## Re: Basic probability question with permutations

I was wondering why it doesn't matter than n >= 4? [Nevermind I understand!]