Example 1B: Using the Fundamental Counting Principle Show 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. How many passwords are possible with 3 letters?There are 15,600 different 3-letter passwords, with no letters repeating, that can be made using the letters a through z. A 3-letter password, with no letters repeating, using the letters a through z, is simply a permutation of 3 letters taken from the alphabet, which has 26 letters.
How many passwords are possible with 5 letters?Therefore the answer is 5 to the fifth power which is 3125. However, if the password is case sensitive, then you actually have 10 choices for each character in your password which would give you 100,000 possibilities.
How many different passwords are possible?Since we cannot reuse whatever we use, this leaves 35 options for the second, 34 for the third, 33 for the fourth, 32 for the fifth, and 31 for the sixth. The total number of possibilities would then be 36 x 35 x 34 x 33 x 32 x 31, which is equal to 1402410240 possible passwords.
How many different passwords are possible with 4 letters?So, we can take any 4 letters of 26 alphabets to form the special type of password and the total number of such passwords is calculated by using permutation. Hence, the required number of total special types of passwords is 358800.
|
