We need to find the number of functions $f: \{1, 2, 3, 4\} \rightarrow \{a, b, c\}$ that are not onto (not surjective).
The total number of possible functions from a set of size $m$ to a set of size $n$ is $n^m$. In this case, $m = 4$ and $n = 3$. Total functions = $3^4$. $ N_{total} = 3^4 = 81 $ There are 81 possible functions from $\{1, 2, 3, 4\}$ to $\{a, b, c\}$.
A function is onto if every element in the codomain is mapped to by at least one element from the domain.
The number of onto functions from a set of size $m$ to a set of size $n$ can be calculated using the principle of inclusion-exclusion:
$ N_{onto} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} (n-k)^m $Substituting $m=4$ and $n=3$:
$ N_{onto} = \sum_{k=0}^{3} (-1)^k \binom{3}{k} (3-k)^4 $Calculating the terms:
Summing the terms:
$ N_{onto} = 81 - 48 + 3 - 0 = 36 $ There are 36 onto functions from $\{1, 2, 3, 4\}$ to $\{a, b, c\}$.The number of functions that are not onto is the total number of functions minus the number of onto functions.
$ N_{\text{not onto}} = N_{total} - N_{onto} $ $ N_{\text{not onto}} = 81 - 36 $ $ N_{\text{not onto}} = 45 $ Therefore, there are 45 functions that are not onto.Let the function $f (x) = \frac{x}{3} + \frac{3}{x} + 3$, $x\neq0$ be strictly increasing in $(-\infty, a_1)\cup(a_2,\infty)$ and strictly decreasing in $(a_3, \alpha_4)\cup(a_4,a_5)$. Then $\sum_{i=1}^5 a_i^2$ is equal to
Let $f(x) = \begin{cases} (x-1)^\frac{1}{2-x}, & x > 1, x \ne 2 \\ k, & x = 2 \end{cases}$ The value of k for which f is continuous at x = 2 is :
Let the function $f (x) = \frac{x}{3} + \frac{3}{x} + 3$, $x\neq0$ be strictly increasing in $(-\infty, a_1)\cup(a_2,\infty)$ and strictly decreasing in $(a_3, \alpha_4)\cup(a_4,a_5)$. Then $\sum_{i=1}^5 a_i^2$ is equal to
Let $f(x) = \begin{cases} (x-1)^\frac{1}{2-x}, & x > 1, x \ne 2 \\ k, & x = 2 \end{cases}$ The value of k for which f is continuous at x = 2 is :