Question

1. how many permutations of a 7-digit phone number exist, assuming that numbers may be repeated and any number 0-9 may be used for any digit?

10!
\\(\frac{10!}{3!}\\)
10^7
70

Explanation:

Step1: Identify the number of choices per digit

Each digit of the 7 - digit phone number can be any of the 10 digits (0 - 9). So, for the first digit, we have 10 choices.

Step2: Account for repetition and use the multiplication principle

Since numbers can be repeated, for the second digit, we also have 10 choices, for the third digit 10 choices, and so on, up to the seventh digit. By the multiplication principle of counting, if we have \(n\) independent events and the number of ways to do each event is \(k_1,k_2,\cdots,k_n\) respectively, the total number of ways to do all the events together is \(k_1\times k_2\times\cdots\times k_n\). Here, \(n = 7\) and \(k_i=10\) for \(i = 1,2,\cdots,7\). So the total number of permutations is \(10\times10\times10\times10\times10\times10\times10=10^{7}\).

Answer:

\(10^{7}\) (corresponding to the option \(10^7\))