Consider the set of all permutations of the numbers 1, 2, 3. If you select a permutationat random, what is the probability that the number 2 is in the middle position?In the first position? Do your answers suggest a simple way of answering the same questions for the set of all permutations of the numbers 1 to 7?

A specific way or order in which a particular set of objects, numbers, or things is arranged is defined as permutation. With help of this concept, the number of arrangements can be determined.

When the numbers 1,2 and 3 are arranged, there are total 6 possible combinations. So, the sample space is expressed as follows.

$123,132, 213, 231, 312, 321$

It is observed that each point of the sample space has an equal probability of $\frac{1}{6}$.

Find the probability that the number 2 is in the middle position by adding the probabilities of the points which has number 2 in the middle.

$\begin{array}{rcl}p& =& \frac{1}{6}+\frac{1}{6}\\ & =& \frac{2}{6}\\ & =& \frac{1}{3}\\ & & \end{array}$

Find the probability that the number 2 is in the first position by adding the probabilities of the points which has number 2 in the middle.

$\begin{array}{rcl}p& =& \frac{1}{6}+\frac{1}{6}\\ & =& \frac{2}{6}\\ & =& \frac{1}{3}\\ & & \end{array}$

Thus, the probability that the number 2 is in the middle position each is and the probability that the number 2 is in the first position each is $\frac{1}{3}$.

It can be observed that when the position of any number is fixed, the probability is reciprocal of total number of digits given. And, thus for numbers 1 to 7, the probability will be $\frac{1}{7}$.