Question

how many one - to - one correspondences are there between two sets with
a. 8 elements each?
b. n elements each? explain your reasoning.
a. there are 40320 one - to - one correspondences between two sets with 8 elements each.
b. there are
one - to - one correspondences between two sets with n elements each. for the first element of the first set, there are
elements of the second set it can be paired with. after that, for the second element of the first set, there are
elements of the second set it can be paired with. the process continues. these numbers of possibilities are
together to get the total number of one - to - one correspondences.
(simplify your answers.)

Explanation:

Step1: Recall the formula for one - to - one correspondences

The number of one - to - one correspondences (bijections) between two sets with \(n\) elements each is given by the factorial of \(n\), denoted as \(n!\). The factorial of a non - negative integer \(n\) is defined as \(n!=n\times(n - 1)\times(n - 2)\times\cdots\times1\). For \(n = 8\), we calculate \(8!\).

Step2: Calculate \(8!\)

We know that \(8!=8\times7\times6\times5\times4\times3\times2\times1\).
First, \(8\times7 = 56\), then \(56\times6=336\), \(336\times5 = 1680\), \(1680\times4=6720\), \(6720\times3 = 20160\), \(20160\times2=40320\), and \(40320\times1 = 40320\).

For the general case with \(n\) elements:

Step1: Understand the process of one - to - one correspondence

For the first element of the first set, we can pair it with \(n\) elements of the second set. After choosing a pair for the first element, for the second element of the first set, we have \(n - 1\) elements left in the second set to pair with (since we need a one - to - one correspondence, we can't reuse the element we already paired). For the third element of the first set, we have \(n-2\) elements left in the second set, and so on.

Step2: Derive the formula for the number of one - to - one correspondences

The total number of one - to - one correspondences between two sets with \(n\) elements each is the product of the number of choices at each step. That is, the number of choices for the first element: \(n\), for the second element: \(n - 1\), for the third element: \(n - 2,\cdots\), for the \(n\)th element: \(1\). So the total number of one - to - one correspondences is \(n\times(n - 1)\times(n - 2)\times\cdots\times1=n!\).

For the blanks in part (b):

• For the first element of the first set, there are \(n\) elements of the second set it can be paired with.
• For the second element of the first set, there are \(n - 1\) elements of the second set it can be paired with.
• The numbers of possibilities are multiplied together. The total number of one - to - one correspondences is \(n\times(n - 1)\times\cdots\times1=n!\)

Answer:

a. The number of one - to - one correspondences between two sets with 8 elements each is \(8!=40320\).
b. For two sets with \(n\) elements each:

• For the first element of the first set, there are \(n\) elements of the second set it can be paired with.
• For the second element of the first set, there are \(n - 1\) elements of the second set it can be paired with.
• The numbers of possibilities are multiplied together to get the total number of one - to - one correspondences, which is \(n!\).