Question
The number of functions $f: \{1, 2, 3, 4\} \rightarrow \{a, b, c\}$, which are not onto, is:
The correct answer is
45
Problem Definition
We need to find the number of functions $f: \{1, 2, 3, 4\} \rightarrow \{a, b, c\}$ that are not onto (not surjective).
- Domain set $A = \{1, 2, 3, 4\}$, so the size of the domain is $|A| = 4$.
- Codomain set $B = \{a, b, c\}$, so the size of the codomain is $|B| = 3$.
Total Number of Functions
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 $\{a, b, c\}$.
Number of Onto Functions
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:
- For $k=0$: $(-1)^0 \binom{3}{0} (3-0)^4 = 1 \cdot 1 \cdot 3^4 = 1 \cdot 81 = 81$
- For $k=1$: $(-1)^1 \binom{3}{1} (3-1)^4 = -1 \cdot 3 \cdot 2^4 = -3 \cdot 16 = -48$
- For $k=2$: $(-1)^2 \binom{3}{2} (3-2)^4 = 1 \cdot 3 \cdot 1^4 = 3 \cdot 1 = 3$
- For $k=3$: $(-1)^3 \binom{3}{3} (3-3)^4 = -1 \cdot 1 \cdot 0^4 = -1 \cdot 0 = 0
Summing the terms:
$ N_{onto} = 81 - 48 + 3 - 0 = 36 $ There are 36 onto functions from $\{1, 2, 3, 4\}$ to $\{a, b, c\}$.Number of Not Onto Functions
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.Was this answer helpful?
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