- #1

Titan97

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## Homework Statement

##A=\{1,2,3,4,5\}##, ##B=\{0,1,2,3,4,5\}##. Find the number of one-one functions ##f:A\rightarrow B## such that ##f(i)\neq i## and ##f(1)\neq 0\text{ or } 1##.

## Homework Equations

Number of derangements of n things = $$n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\cdots+(-1)^n\frac{1}{n!}\right)$$

## The Attempt at a Solution

[/B]

If either of ##0## and ##1## are not included in the range of ##f##, then total possibilities = ##2\times 44=88##

If both ##0,1## are included in the range of ##f##, then one number from ##\{2,3,4,5\}## has to be mapped to ##1##. Number of possibilities is ##\binom{4}{1}##. Let me map ##1## to ##5##.

For the remaining ##5## numbers, one number other than ##0,1## has to be excluded (since ##0,1## are in the range of ##f##) Number of such possibilities is ##\binom{3}{1}##. Let the number removed be ##4##.

The remaining numbers ##\{0,1,2,3\}## has to be mapped to ##\{2,3,4,5\}##.

**How can I use principle of inclusion and exclusion here?**