Question

1. Some automobiles have six-button locks. To gain entry, push each of the six buttons, in a particular order, to unlock the door. How many different five-digit codes are possible?

2. the "Pick 3" at horse racetracks requires that a person select the winning horse for three consecutive races. If the first race has seven entries, the second race eleven entries, and the third race eight entries, how many different possible tickets might be purchased?


3. If a state issued license plates using the scheme of four letters followed by two digits, how many plates could it issue?


4. In how many ways could a club of 18 members choose a president, vice president, and treasurer?


5. In how many ways could a club of 28 appoint a committee of 4 people?

Answer 1

A.720 different five-digit codes are possible with a six-button lock.

2. 616 different possible tickets might be purchased for the "Pick 3" at horse racetracks.

3.A state could issue 456,976,000 license plates with four letters followed by two digits.

4.There are 5,832 ways to choose a president, vice president, and treasurer from a club of 18 members.

5.A club of 28 can appoint a committee of 4 people in 17,154 different ways.

Explanation:

To find the number of different five-digit codes possible with a six-button lock, you're essentially looking for the number of permutations of 6 items taken 5 at a time because you can choose each button in a particular order. The formula for permutations is:

P(n, k) = n! / (n - k)!

In this case, n = 6 (number of buttons) and k = 5 (number of choices). So,

P(6, 5) = 6! / (6 - 5)!

P(6, 5) = 6! / 1

P(6, 5) = 6 * 5 * 4 * 3 * 2 = 720

There are 720 different five-digit codes possible.

To find the number of different possible tickets that can be purchased for the "Pick 3" at horse racetracks, you need to multiply the number of choices for each race. The total number of possibilities is:

7 (choices for the first race) * 11 (choices for the second race) * 8 (choices for the third race) = 616 different possible tickets.

For state-issued license plates with four letters followed by two digits, you need to consider the number of choices for each part:

For the four letters, there are 26 choices for each letter (assuming the alphabet), so there are 26^4 possibilities.

For the two digits, there are 10 choices for each digit (0-9), so there are 10^2 possibilities.

Therefore, the total number of license plates that can be issued is 26^4 * 10^2 = 456,976,000 plates.

To choose a president, vice president, and treasurer from a club of 18 members, you can use permutations because the order in which the positions are filled matters.

P(18, 3) = 18! / (18 - 3)!

P(18, 3) = 18! / 15!

P(18, 3) = 18 * 17 * 16 = 5,832

There are 5,832 different ways to choose these positions.

To appoint a committee of 4 people from a club of 28, you can use combinations because the order doesn't matter.

C(28, 4) = 28! / (4! * (28 - 4)!)

C(28, 4) = 28! / (4! * 24!)

C(28, 4) = (28 * 27 * 26 * 25) / (4 * 3 * 2 * 1) = 17,154

There are 17,154 different ways to appoint the committee.