Thread: 123 written in a random order probability any are in the right place

i have the question if 1,2 & 3 are written in a random order whats the chance one or more of the didgits are in the right place

i know the probability is 2/3 as i can easily work it out in my head but i have no idea how to work it out mathematically ive tryed loads of different ways but alway end up with something greater than 1

There are 3! possible ways to arrange the numbers.

There are (2)2! ways that at least one letter will be in the correct position.

(when one number is held fixed there are 2! ways to arrange the other numbers)

2!2!/3! = 4/6 = 2/3

Hello, renlok!

Kivit got the right answer, but I don't understand his reasoning.

If 1,2 & 3 are written in a random order
what's the chance one or more of the digits are in the right place?

There are possible permutations.


There are 3 ways that exactly one digit is in the right place: .132, 321, 213

There is no way that exactly two digits are in the right place.

There is 1 way that all three digits are in the right place: 123

Hence, there are 4 ways that at least one digit is in the right place.


The probability is: .


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Here's another approach . . . a rather primitive one.

There are possible permutations.

How many of them have no digits in the right place?


The first digit cannot be "1"; it must be "2" or "3".

If the first digit is "2": . 2 _ _
. . we must place the "1" and "3".
The "3" cannot be the third digit; it must be the second digit.
. . We have: .231

If the first digit is "3": .3 _ _
. . we must place the "1" and "2".
The "2" cannot be the second digit; it must be the third digit.
. . We have: .312

There are only two ways that no digits are in the right place.

Hence, there are: . ways with at least one digit correct.

is there no way to do it with nPr and nCr becausee the next question is the same but with n numbers

(n-1)(n-1)!/n!