Answer
a) $a_{n}$ = $n\times a_{n-1}$
b) $a_{n}$ = n!
Work Step by Step
a)
Let $a_{n}$ be the number of permutations of a set with n elements.
Therefore, the number of permutations of a set with n-1 elements is $a_{n-1}$.
Let us consider all $a_{n-1}$ permutations of the set with n-1 elements. For each permutation, we can put an extra element at one of the n (= 1(beginning)+(n-2)(spaces) + 1(end) ) places.
So, for each of the a_{n-1} permutations of the set, we have n permutations of the set with n elements.
Thus, $ a_{n} = n \times a_{n-1} $
b)
$ a_{n} = n \times a_{n-1} $
$ a_{n-1} = (n-1) \times a_{n-2} $
$ a_{n-2} = (n-2) \times a_{n-3} $
......................
$ a_{1} = 1 \times a_{0} $
where $a_0$ = no. of ways of arranging zero things = 1
Multiplying all these equations and canceling similar terms gives,
$ a_{n}$ = $n \times (n-1) \times (n-2) $ . . . $2\times 1 \times a_0$
Putting a_0 = 1 gives,
$ a_{n}$ = $n \times (n-1) \times (n-2) $ . . . $2\times 1$
which is also written as n! ( n-factorial ) in maths.
