Understanding Integer Permutations: X Natural Positions Explained

[annoymage]

Homework Statement

Given a random permutation of integer in the set (1,2,3,4,5), let X equal the number of integer that are in their natural position.

The Attempt at a Solution

i don't understand "number of integer that are in their natural position"
can someone explain it. please

 

[Dick]

It's the number of integers that are mapped to themselves under the random permutation.

 

[annoymage]

hmm still confused, I am sorry for my bad english,

let me tell what i understand,

the universal set are

(1,2,3,4,5),(5,4,3,2,1),(1,2,3,5,4) and so on,
there's 5!=120 in total,

so, which are integer that maps to itself? can give me some example.. :)

 

[swuster]

If, for example, your permutation is (2,3,1,4,5) then X = 2 because 4 and 5 are in their original, or natural positions.

 

[annoymage]

OOOOOOOOOOOOOOOOOOOOOOOO, i get it, thankssss,
thank you very much

 

[snshusat161]

If, for example, your permutation is (2,3,1,4,5) then X = 2 because 4 and 5 are in their original, or natural positions.

But what about 3 and 1

 

[annoymage]

he means, that, 4 and 5 is the integer in its natural position...
so, x=2 which means, there are two integer in its natural position..

(2,3,1,*,*) is not in their natural position because (1,2,3,*,*) is their actual position

example:

(5,2,3,4,1) , so 2,3, and 4 are in their natural position, so x=3

 

[snshusat161]

Thank, I got it.