I came across this question when revising:
The numbers 1 to 11 are arranged in a line (uniformly at random). What is the probability that the first number and last number are both even?
My method
Even numbers: 2,4,6,8,10
Odd numbers: 1,3,5,7,9,11
I was thinking since the probability for the first number to be even would be $\frac{5}{11}$ and the probability for the last number to be even would be $\frac{4}{10}$(since if the first number is already taken and is even, so there are 10 numbers left and 4 of them are even)
But upon rethinking, I started to have doubts about my method as I realised that the sequence in which the number can be placed must also be taken into account.
Does anyone know how to solve this, any help is truly appreciated. Thank you so much in advance.
Best Answer
There are $\color{magenta}{11\times10\times9\times\cdots\times1}=\color{magenta}{11!}$ ways of permuting these numbers. Out of them, the first and the last can be chosen even in $5\times4=20$ ways so that the total number of ways in which the first and the last are even is $5\times\color{magenta}{9\times8\times7\times\cdots\times1}\times4= 20\times\color{magenta}{9!}$.
Consequently, the required probability is $$\frac{20\times 9!}{11!}=\frac{2}{11}$$