Consider two sets $A = \{2, 3, 5, 7, 11, 13\}$ and $B = \{1, 8, 27\}$. Let $f$ be a function from $A$ to $B$ such that for every element $b$ in $B$, there is at least one element $a$ in $A$ such that $f(a) = b$. Then, the total number of such functions $f$ is

We need to find functions where every element in $B$ gets mapped to by at least one element from $A$. This is called an onto function.

$A = \{2, 3, 5, 7, 11, 13\}$, so $|A| = 6$

$B = \{1, 8, 27\}$, so $|B| = 3$

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We'll use the Inclusion-Exclusion Principle: count all functions, then subtract the bad ones (those that miss elements).

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Each of the $6$ elements in $A$ can map to any of the $3$ elements in $B$.

Total functions $= 3^6 = 729$

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Functions that miss element $1$:

Elements in $A$ can only map to $\{8, 27\}$

Count $= 2^6 = 64$

Functions that miss element $8$:

Elements in $A$ can only map to $\{1, 27\}$

Count $= 2^6 = 64$

Functions that miss element $27$:

Elements in $A$ can only map to $\{1, 8\}$

Count $= 2^6 = 64$

Functions missing at least one element $= 3 \times 64 = 192$

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We overcounted functions that miss two elements, so we add them back.

Functions that miss both $1$ and $8$:

Elements in $A$ can only map to $\{27\}$

Count $= 1^6 = 1$

Functions that miss both $1$ and $27$:

Elements in $A$ can only map to $\{8\}$

Count $= 1^6 = 1$

Functions that miss both $8$ and $27$:

Elements in $A$ can only map to $\{1\}$

Count $= 1^6 = 1$

Functions missing exactly two elements $= 3 \times 1 = 3$

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Functions missing all three elements $= 0$ (impossible)

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Using Inclusion-Exclusion:

Onto functions $= 729 - 192 + 3 - 0 = 540$

Therefore, the total number of onto functions from $A$ to $B$ is $540$.