Example 1B: Using the Fundamental Counting Principle
A password for a site consists of 4 digits followed by 2 letters. The letters A and Z are not used, and each digit or letter many be used more than once. How many unique passwords are possible?
digit digit digit digit letter letter
10 × 10 × 10 × 10 × 24 × 24 = 5,760,000
There are 5,760,000 possible passwords.
Holt McDougal Algebra 2
Permutations and Combinations
Check It Out! Example 1a
A “make-your-own-adventure” story lets you choose 6 starting points, gives 4 plot choices, and then has 5 possible endings. How many adventures are there?
number of starting points
×
number
of possible endings
×
=
6 × 4 × 5 = 120
There are 120 adventures.
Holt McDougal Algebra 2
Permutations and Combinations
Check It Out! Example 1b
A password is 4 letters followed by 1 digit. Uppercase letters (A) and lowercase letters (a) may be used and are considered different. How many passwords are possible?
Since both upper and lower case letters can be used, there are 52 possible letter choices.
letter letter letter letter number
52 × 52 × 52 × 52 × 10 = 73,116,160
There are 73,116,160 possible passwords.
Holt McDougal Algebra 2
Permutations and Combinations
Example 2B: Finding Permutations
How many ways can a stylist arrange 5 of 8 vases from left to right in a store display?
Divide out common factors.
= 8 • 7 • 6 • 5 • 4
= 6720
There are 6720 ways that the vases can be arranged.
Holt McDougal Algebra 2
Permutations and Combinations
Check It Out! Example 2a
Awards are given out at a costume party. How many ways can “most creative,” “silliest,” and “best” costume be awarded to 8 contestants if no one gets more than one award?
= 8 • 7 • 6
= 336
There are 336 ways to arrange the awards.
Holt McDougal Algebra 2
Permutations and Combinations
Check It Out! Example 2b
How many ways can a 2-digit number be formed by using only the digits 5–9 and by each digit being used only once?
= 5 • 4
= 20
There are 20 ways for the numbers to be formed.
Holt McDougal Algebra 2
Permutations and Combinations
Lesson Quiz
1. Six different books will be displayed in the library window. How many different arrangements are there?
2. The code for a lock consists of 5 digits. The last number cannot be 0 or 1. How many different codes are possible?
80,000
720
3. The three best essays in a contest will receive gold, silver, and bronze stars. There are 10 essays. In how many ways can the prizes be awarded?
4. In a talent show, the top 3 performers of 15 will advance to the next round. In how many ways can this be done?
455
720
Holt McDougal Algebra 2
Permutations and Combinations
I believe you are correct, and that your lecturer's solution overcounts the solution by a factor of two exactly.
By looking at the formulas provided, it seems that your reasoning is as follows: choose two of the six spaces to be letters, choose the letters used, then choose the numbers used in the remaining four spaces. Hence, $\binom{6}{2} \cdot 26^2 \cdot 10 \cdot 9 \cdot 8 \cdot 7$.
On the other hand, your lecturer's reasoning seems to be: choose which letters and numbers to use first, then permute the characters arbitrarily. Hence, $26^2 \cdot \binom{10}{4} \cdot 6!$. However, the problem with this approach is that it overcounts the number of distinct ways you can permute the letters. For example, if two $a$'s were chosen for the letters (denote then $a_1$ and $a_2$ to distinguish them from each other) and, say, $1234$ were chosen for the digits, the passwords $$a_1a_21234 \quad \text{and} \quad a_2a_11234$$ would be counted differently, even though they're the same password. This doesn't happen in your case, since fixing the positions of the letters first before choosing them results in each pair of letters leading to a different password combination.