Problem on combination - calculating the number of possible area codes

Question

For years, telephone area codes in the USA and Canada consisted of a sequence of three digits. The first digit was an integer between 2 and 9, the second digit was either 0 or 1, and the third digit was any integer from 1 to 9. How many area codes were possible? How many area codes starting with a 4 were possible?

Can someone help me with this problem, please? And could you tell me how to understand and a proceed quite easily with combinatorial problems like these? I'm having a hard time solving even a couple of problems. Thank you in advance! :)

Answer 1

Apply the rule of product which can be paraphrased for counting problems to be the following:

If you wish to count how many outcomes exist for a scenario and you can describe each outcome uniquely via a sequence of steps with choices so that each outcome is described by exactly one sequence of choices and the number of options available at each step does not depend on the previously made selections (though the options themselves may change) then the total number of outcomes is the product of the number of options available at each step.

This is just formalizing what you should already know about areas of grids and such and extending to higher dimensions and giving it combinatorial flavor and context. The number of ways you could have a letter from $\{A,B,C,D\}$ followed by a number from $\{1,2,3\}$ would be $4\times 3=12$, which you should be able to verify by looking at the grid:

$$\begin{array}{r|ccc}&1&2&3\\\hline A&A1&A2&A3\\B&B1&B2&B3\\C&C1&C2&C3\\D&D1&D2&D3\end{array}$$

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For your problem, we break it up via rule of product by defining the steps to be:

• Pick the first digit (It may be any integer between 2 and 9, so there are eight options)
• Pick the second digit (It must be a 0 or a 1, so there are two options)
• Pick the third digit (It may be any integer between 1 and 9, so there are nine options)

Multiplying the number of options for each step gives the result.

For the second part of the problem, the first step will have a different number of options available.

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As an additional example to highlight that the options themselves can change, the number of permutations of $1234$ (1234,1243,1324,1342,1423,1432,...) will be $4!=4\times 3\times 2\times 1=24$. This is seen by the following steps:

• Pick the first digit (It can be any of $1,2,3,4$, so there are four options)
• Pick the second digit (It can by any of $1,2,3,4$ except whatever was selected during step 1, so there are only three remaining options)
• Pick the third digit (It can be any of $1,2,3,4$ except whatever was selected during either of steps 1 or 2, so there are only two remaining options)
• Pick the fourth digit (It must be the only remaining unused digit, so there is only one option)

which, when multiplying the number of options available at each step gives what should be a familiar answer of $4\times 3\times 2\times 1=4!$

Answer 2

$8\times2\times9 = 144$.

There are $8$ spots for the first option, $2$ spots for the second option, and $9$ spots for the third option.

Now try to do it with using 4 as the starting digit.