If n men and n women meet in order to marry then there are exactly n! different arrangements such that each man is married with exactly one woman and vice versa.
Hint: Give numbers to the men and women from 1 to n and denote an arrangement as n- tuple where position i stores number j if the man with number i is married with woman with number j. For the inductive step, focus on George and consider how many possibilities exist for George to marry a woman. Then combine this with the arrangements for the remaining men and women.
Any help will be appreciate-able.
I took the n tuple as
f(n) = (1, 2, 3, 4, ......... n)
but can i use the induction as
n! or 1.2.3 .....n.(n+1)!
I do not understand that how i will build the relation.
What a pain it is to prove this by induction. Numbering the brides and grooms, and then giving an ordering of the grooms by bride, seems much more straightforward.
I assume you already have the basis step. The induction step can be accounted for as follows. Suppose you already have $k!$ arrangements for $n = k$; in these arrangements, let the brides and grooms be denoted by $\{b_i\}, \{g_i\}, i = 1, 2, \ldots$.
Given any one of the $k!$ arrangements of the $n = k$ brides and grooms, we can introduce $b_{k+1}$ and $g_{k+1}$ in $k+1$ ways. One way is for the two to marry each other. Then there are $k$ different ways for both of them to marry one of the first $k$ brides and grooms. (You need to figure out how to show this.)
Since this reasoning maps each arrangement of $k$ brides and grooms to $k+1$ arrangements of $k+1$ brides and grooms, and the arrangements are distinct (you need to figure out how to show this, too), there must be $k+1$ times as many arrangements of $k+1$ brides and grooms as there are of $k$ brides and grooms, and the induction step is shown.