Counting One-to-One Functions

The problem requires finding the total number of one-to-one functions that can be defined from a set A = {1, 2, 3} to a set B = {1, 2, 3, 4, 5}. A function $f: A \to B$ is defined as one-to-one (or injective) if distinct elements in the domain A map to distinct elements in the codomain B. Mathematically, this means that for any $x_1, x_2 \in A$, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.

The domain set is A = {1, 2, 3}, so it has $n(A) = 3$ elements.

The codomain set is B = {1, 2, 3, 4, 5}, so it has $n(B) = 5$ elements.

Since the number of elements in the domain ($n(A)=3$) is less than the number of elements in the codomain ($n(B)=5$), it is possible to construct one-to-one functions.

Permutations Principle for Functions

To count the number of one-to-one functions, we can consider the choices for mapping each element from the domain A to the codomain B:

• The first element of A (1) can be mapped to any of the 5 elements in B. Thus, there are 5 possible choices.
• The second element of A (2) must be mapped to an element in B that is different from the one chosen for the first element. This leaves $5-1 = 4$ possible choices.
• The third element of A (3) must be mapped to an element in B that is different from the two elements already chosen for the first and second elements. This leaves $5-2 = 3$ possible choices.

The total number of distinct one-to-one functions is the product of the number of choices available for each element in the domain. This scenario corresponds to the concept of permutations, specifically, the number of ways to arrange $n$ items selected from a set of $m$ distinct items, denoted as $P(m, n)$ or $^mP_n$. The formula for permutations is:

$$ P(m, n) = \frac{m!}{(m-n)!} $$

Here, $m$ represents the total number of items to choose from (the size of the codomain), and $n$ represents the number of items to choose and arrange (the size of the domain).

Calculation Steps for $P(5, 3)$

For this specific problem:

• The size of the codomain is $m = n(B) = 5$.
• The size of the domain is $n = n(A) = 3$.

We apply the permutation formula $P(m, n)$:

$$ P(5, 3) = \frac{5!}{(5-3)!} $$

First, calculate the value in the denominator:

$$ P(5, 3) = \frac{5!}{2!} $$

Next, compute the factorials:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$2! = 2 \times 1 = 2$

Now substitute these values back into the formula:

$$ P(5, 3) = \frac{120}{2} $$

Performing the division gives the result:

$$ P(5, 3) = 60 $$

Result of One-to-One Function Count

Thus, there are exactly 60 distinct one-to-one functions that can be formed from the set {1, 2, 3} to the set {1, 2, 3, 4, 5}. This matches the calculated value.