Question

Consider two sets A = [1, 2, 3] and B = [1, 2, 3, 4, 5]. What is the total number of one-to-one functions from A to B?

• A. 125
• B. 243
• C. 60
• D. 30

Answer 1

Answer: (C)

60

Answer 2

The problem asks to find the total number of one-to-one functions from set A to set B.

Given sets:

A = [1, 2, 3] B = [1, 2, 3, 4, 5]

First, determine the number of elements in each set:

Number of elements in set A, n(A) = 3 Number of elements in set B, n(B) = 5

A function f: A → B is one-to-one (or injective) if every distinct element in A maps to a distinct element in B. This means that for any x₁, x₂ ∈ A, if x₁ ≠ x₂, then f(x₁) ≠ f(x₂).

To construct a one-to-one function from A to B:

1. The first element of A (say, 1) can be mapped to any of the 5 elements in B. (5 choices)
2. The second element of A (say, 2) must be mapped to one of the remaining (5 - 1) = 4 elements in B, as it must be distinct from the image of the first element. (4 choices)
3. The third element of A (say, 3) must be mapped to one of the remaining (5 - 2) = 3 elements in B, as it must be distinct from the images of the first two elements. (3 choices)

The total number of ways to define such a function is the product of the number of choices for each element:

Total number of one-to-one functions = 5 × 4 × 3 = 60.

Alternatively, this can be solved using the permutation formula. The number of one-to-one functions from a set with 'k' elements to a set with 'n' elements (where k ≤ n) is given by the number of permutations of 'n' items taken 'k' at a time, denoted as P(n, k) or ⁿPₖ.

Here, k = n(A) = 3 and n = n(B) = 5.

So, the number of one-to-one functions = P(5, 3)

P(n, k) = n! / (n - k)!

P(5, 3) = 5! / (5 - 3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 5 × 4 × 3 = 60