If $A,B = \left\{1,2,3,4,5,6\right\}$ and $f(x)$ be a onto function which is defined from $A$ to $B$ such that
$f(1) = 2$ and $f(i) \neq i \;\; \forall \; i\in \{1,2,3,4,5,6\}$.
$\bf{My\; Try::}$ No. of onto function from $A$ to $B$ is equivalent to put $6$ letters into $6$ envelopes So that
letter so that no Letter goes to its envelope and Letter with Lebel $(1)$ goes to envelope with Lebel $(2)$
Now I did not understand how can i solve it.
Help Required
Thanks.
You can count all these functions by inclusion/exclusion. Let $\cal U$ be the set of all functions $f$ from $A=\{1,\ldots,6\}$ onto $B=\{1,\ldots,6\}$ with the property that $f(1)=2$. As $f$ is to be onto, other values of $f(k)$ are obtained by writing down the numbers $1,3,4,5,6$ in some order: there are $5!$ possibilities.
Note that since $A$ and $B$ are of the same size, every onto function is one-to-one as well.
Now for $k=1,\ldots,6$, write $$F_k=\{f\in{\cal U}\mid f(k)=k\}\ .$$ Obviously $F_1=\varnothing$; since functions in $\cal U$ are one-to-one, $F_2=\varnothing$ too. So the number of functions we want to count is $$|\overline{F_3}\cap\cdots\cap\overline{F_6}| =|{\cal U}|-|F_3|-\cdots+|F_3\cap F_4|+\cdots\ .$$ The functions in $F_3$ satisfy $f(1)=2$, $f(3)=3$, other $f(k)=1,4,5,6$ in some order; so $|F_3|=4!$. Doing other terms in the same way gives the required number as $$5!-4\times4!+6\times3!-4\times2!+1!\ .$$
Hint: The answer is $ \frac{ D_n}{n-1}$.
For each $f \in D_n$, let $ f \in F_i$ if $ f(1) = i$. Show that $|F_1| = 0$ and that $|F_i|= |F_j| $ for $ i, j \neq 1$.
[Given the terminology that you used, I'm assuming that you are familiar with Derangements.]
