Step-by-step Solution:

The number of onto (surjective) functions from set \( A \) to set \( B \), where \( |A| = 6 \) and \( |B| = 3 \), is given by the formula: \[ S(m, n) = n^m - \sum_{k=1}^{n} \binom{n}{k} (n-k)^m (-1)^k \] where:
\( m = 6 \) (number of elements in \( A \))
\( n = 3 \) (number of elements in \( B \))
This formula is derived from the principle of inclusion-exclusion.
Step 1: Compute Total Functions
The total number of functions from \( A \) to \( B \) (without restriction) is: \[ 3^6 = 729 \] Step 2: Exclude Non-Surjective Cases
We subtract cases where one or more elements of \( B \) are missing:
1. At least one element of \( B \) is missing:
Choose one element to exclude: \( \binom{3}{1} = 3 \).
Remaining elements receive values: \( 2^6 = 64 \).
Contribution: \( 3 \times 64 = 192 \).
2. At least two elements of \( B \) are missing:
Choose two elements to exclude: \( \binom{3}{2} = 3 \).
Remaining element receives values: \( 1^6 = 1 \).
Contribution: \( 3 \times 1 = 3 \).
Step 3: Apply Inclusion-Exclusion \[ S(6,3) = 729 - 192 + 3 = 540 \] Final Answer: \( \mathbf{540} \)
So, the number of onto (surjective) functions from \( A \) to \( B \) is 540.