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Question: $F$ is the set of all functions from a $n$-set to $B=\{a,b,c\}$. In a random experiment of selection of functions from $f$ assume that every $f\in F$ is equally likely. What is the probability that such a function has $a$ in its image?

I know the problem can be easily tackled by calculating the probability of selecting a function not having $a$ in its image ($=2^n/3^n$), and finding the desired probability by subtracting it from $1 (=\frac{3^n-2^n}{3^n})$. However I wanted to see if I could solve it using the Inclusion-Exclusion Principle too:

The total number of functions $f: \{a_1,a_2,a_3...a_n\} \to \{a,b,c\}$ would be $3^n$. Let $P_i$ be the property "$a_i$ doesnt map to $a$". Hence,

$$\sum_{1\leq i\leq n}|P_i| = \sum 2 \cdot 3^{n-1} = n\cdot2\cdot 3^{n-1}$$ (Choose from $\{b,c\}$ for $a_i$, $\{a,b,c\}$ for the rest)

$$\sum_{1\leq i \leq j \leq n}|P_i \cap P_j| = \sum 2^2 \cdot3^{n-2}n={n \choose 2}2^2\cdot3^{n-2}$$ (Choose from $\{b,c\}$ for ${n \choose 2}$ pairs of $(a_i, a_j)$)

Hence, $$\sum\left|\bigcap_{i=1}^{r} P_i\right|={n \choose r}2^r\cdot3^{n-r}$$ (Choose from $\{b,c\}$ for ${n\choose r}$ selections of $(a_i,a_j...a_{i+r})$)

By IEP, Number of functions which map having $a$ in its image= $$\left|\bigcap_{i=1}^n\overline{P_i}\right|=\left|\overline{\bigcup_{i=1}^n P_i}\right|=3^n -\left|\bigcup_{i=1}^n P_i\right|=3^n-2\cdot3^{n-1}n+{n \choose 2}2^2\cdot3^{n-2}\cdots+(-1)^n2^n = (3-2)^n=1$$ The number of functions which have $a$ in their image definitely isn't $1$. Is there something wrong with this approach?

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A good careful calculation, unfortunately of the wrong thing.

The Inclusion/Exclusion argument in the OP shows, in a correct but lengthy way, that there is exactly one function that maps all the $a_i$ to $a$.

For the union of the $P_i$ is the set of all functions that miss $a$ somewhere. Subtracting the cardinality of the union from $3^n$ counts the functions that miss $a$ nowhere. There is only one such function.

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  • $\begingroup$ So, is it possible to find the number of functions using IEP (maybe by not subtracting the union from $3^n$)? $\endgroup$ – Vibhav Pant Jan 3 '15 at 5:05
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    $\begingroup$ An Inclusion/Exclusion argument very similar to yours will work, though as you know it is not optimal. We count the functions that map $1$ to $a$, $2$ to $a$ and so on and add up. From these we must subtract the functions that send both $1$ and $2$ to $a$, and so on for all pairs. And add back $\dots$. When we divide by $3^n$ we get $1$ minus the binomial expansion of $\left(1-\frac{1}{3}\right)^n$. $\endgroup$ – André Nicolas Jan 3 '15 at 7:17

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