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 $\begingroup$ I'm having a little trouble with an exercise, due to a flawed logic that I need to fix. I apologize in advance if anything of this is silly, or for my language as I'm not a native English speaker. I have an alphabet of 21 letters and 10 digits (0-9) and I need to calculate the probability of creating a password that has a total length of 8 characters, with exactly 2 digits and 6 letters. First, I defined the total of outcomes, which are $31^8$: I've imagined 8 boxes, each one can have 31 (the total length of the alphabet) possible characters. Now, for the likely outcomes I've imagined to calculate all the ways to take 2 digits out of 10 ($\binom{10}{2}$) and multiply them for the possible ways to take 6 letters from 21 total letters($\binom{21}{6}$). I thought of combinations because no order is apparently required. The solution is different on these outcomes: I was right on $31^8$ but the likely outcomes are calculated as $\binom{8}{2}*10^2*21^6$ for a result of $P(E) = \frac{\binom{8}{2}*10^2*21^6}{31^8}$. I can't understand this solution, nor what I'm doing wrong. Can anybody help me out? asked Mar 17, 2018 at 18:09
$\endgroup$ $\begingroup$ Let's think about how you obtained $31^8$ possible outcomes. You realized that there were $31$ choices for each of the eight positions in the password, including $21$ letters and $10$ digits. Since there are eight positions, there are $\binom{8}{2}$ ways to choose the positions of the digits. Each of these two positions can be filled with a digit in $10$ ways, giving us the factor $10^2$. Each of the remaining six positions can be filled with a letter in $21$ ways, giving us the factor $21^6$. Hence, the number of favorable cases is $$\binom{8}{2}10^221^6$$ What are you doing wrong? The number $\binom{10}{2}$ counts the number of ways of selecting two of the ten digits. Therefore, you have selected two different digits, but digits may be repeated. Similarly, $\binom{21}{6}$ counts the number of ways of selecting six different letters, but letters may be repeated. You also did not choose the positions of the letters. answered Mar 17, 2018 at 18:14
N. F. TaussigN. F. Taussig 69.1k13 gold badges52 silver badges70 bronze badges $\endgroup$ 1 $\begingroup$ Here you should not do $\binom{10}{2}$ because you can make passwords with 2 same digits in it,like 22,33 . Here you just have to choose number of places for digits among 8 given places, $\binom82$ ways. Now since you can place same digits and letters, say for 2 places for letters you can choose first letter in 10 way and next letter also 10 way, thats why $10^2$ and the same with letters. answered Mar 17, 2018 at 18:13
kayushkayush 2,39111 silver badges26 bronze badges $\endgroup$ How many passwords can you create if the password must consist of 2 letters followed by 4 digits with repeats allowed?Since both upper and lower case letters can be used, there are 52 possible letter choices. There are 73,116,160 possible passwords.
How many passwords are possible?An eight-character password using only lowercase and uppercase characters has 200 billion possible combinations.
How many passwords are possible with 4 digits and 2 letters?The two letters can be chosen without repeating in 26*25 = 650 ways. There are 10 digits between 0 and 9. Any four can be selected without repeating in 10*9*8*7 = 5040 ways. The combination of a letters and 4 digits can be made in 650*5040 = 3,276,000 ways.
How many unique passwords can be made from 2 letters and 3 digits?=26×26×10×10×10=676000.
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