How many ways can an IRS auditor select 3 of 11 tax returns for an audit?

There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?

License plates are made using 3 letters followed by 3 digits. How many plates can be made if repetition of letters and digits is allowed.

At the first tri-city meeting, there are 8 people from town A, 7 people from town B, and 5 people from town C. If a council consisting of 5 people is randoml selected, find the probabilit that 3 are from town A and 2 are from town B.

The auditor chooses the first return out of 11, that makes 11 choices. For the second, she is left with 10, so there are 11*10 ways. For the third, she has 9 returns left from which to choose, so there are 11*10*9=990 ways to choosing the first 3 returns.

However, out of the 990 ways to choose the returns, each combination (say A,B,C) would have been repeated in 6 different orders, like ABC, BAC, CAB, BCA, ACB, CBA. The number of repetitions can be calculated by 6=3!, where three is the number of returns chosen.

Thus the number of ways she can choose 3 returns would be
11*10*9/3!=990/6=165.

It's your turn to choose the sub-committees and try the rest of the exercises.

If you have difficulties, post what you have or what the difficulties are.

How many ways can an IRS auditor select 3 of 11 tax returns for an audit?

Well, let's see, selecting tax returns for an audit, huh? That's like trying to pick your least favorite vegetable at the grocery store. But back to the question -- to find the number of ways, we can use the combination formula. It's like figuring out the number of ways you can arrange your sock drawer, except it's much less exciting. So, using the combination formula, we get 11 choose 3, which is equal to 165. So there are 165 ways for the IRS auditor to spoil someone's day.

There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?
Forming a subcommittee? Oh, that sounds like picking a dinner menu with your friends who can never agree on what to eat. Nevertheless, let's solve this. We just need to find the number of ways to choose 3 members out of 8, which is 8 choose 3. And after crunching the numbers, we get 56. So, there are 56 different subcommittees possible. Just make sure they can all agree on what to order for lunch!

License plates are made using 3 letters followed by 3 digits. How many plates can be made if repetition of letters and digits is allowed?
Ah, license plates, the perfect way to express your creativity within the guidelines of the law. So, we have 26 letters in the English alphabet and 10 digits from 0 to 9. Since repetition is allowed, it's like having a never-ending supply of bad jokes. So, we have 26 possibilities for each letter, and 10 possibilities for each digit. Multiply them together, and we get 26^3 * 10^3, which is 17,576,000. That's a whole lot of plates! But remember, even with a fancy plate, you can't escape traffic jams.

At the first tri-city meeting, there are 8 people from town A, 7 people from town B, and 5 people from town C. If a council consisting of 5 people is randomly selected, find the probability that 3 are from town A and 2 are from town B.
Sounds like a council meeting where everyone argues about whose town pizza is the best. Alright, let's calculate the probability. The total number of ways to select 5 people from the total pool of 20 is 20 choose 5. And the number of ways to select 3 people from town A is 8 choose 3, while the number of ways to select 2 people from town B is 7 choose 2. To find the probability, we just need to divide these two numbers. So, (8 choose 3) * (7 choose 2) / (20 choose 5) gives us approximately 0.0787. In other words, there's about a 7.87% chance that town A will dominate the council, and town B will have to settle for making the best pizza.

To calculate the number of ways to select 3 tax returns for an audit, you would use combinations. The formula for combinations is:

C(n, r) = n! / (r!(n-r)!)

In this case, n = 11 (number of tax returns) and r = 3 (number of returns to be audited).
So, the number of ways to select 3 tax returns for an audit is:

C(11, 3) = 11! / (3!(11-3)!)
= 11! / (3! * 8!)
= (11 * 10 * 9) / (3 * 2 * 1)
= 165

Therefore, there are 165 different ways an IRS auditor can select 3 of the 11 tax returns for an audit.

For the subcommittee of 3 members from a board of directors with 8 members, we can use combinations as well.

C(8, 3) = 8! / (3!(8-3)!)
= 8! / (3! * 5!)
= (8 * 7 * 6) / (3 * 2 * 1)
= 56

There are 56 different subcommittees possible with 3 members selected from a board of 8 members.

For the license plates with 3 letters followed by 3 digits, where repetition of letters and digits is allowed, we can use the multiplication principle.

Since there are 26 letters (A-Z) and 10 digits (0-9), the number of possible license plates can be calculated as:

26 * 26 * 26 * 10 * 10 * 10
= 26^3 * 10^3
= 17,576,000

Therefore, there are 17,576,000 possible license plates that can be made if repetition of letters and digits is allowed.

To find the probability that 3 people are from town A and 2 people are from town B, we need to calculate the number of ways this can occur, and then divide by the total number of possible combinations.

The number of ways to choose 3 people from town A out of 8, and 2 people from town B out of 7, can be calculated as:

C(8, 3) * C(7, 2) = (8! / (3! * (8-3)!) ) * (7! / (2! * (7-2)!) )
= (8! / (3! * 5!) ) * (7! / (2! * 5!) )
= (8 * 7 * 6) / (3 * 2 * 1) * (7 * 6) / (2 * 1)
= 56 * 21
= 1,176

The total number of ways to choose a council of 5 people from a group of 20 (8 from town A, 7 from town B, and 5 from town C) can be calculated as:

C(20, 5) = 20! / (5!(20-5)!)
= 20! / (5! * 15!)
= (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1)
= 15,504

Therefore, the probability that 3 people are from town A and 2 people are from town B can be calculated as:

P(3 from town A and 2 from town B) = number of ways to choose 3 from town A and 2 from town B / total number of ways to choose a council of 5 people

P(3 from town A and 2 from town B) = 1,176 / 15,504 ≈ 0.076

To find the number of ways to select 3 of 11 tax returns for an audit, you can use the combination formula. The formula for combination is:

nCr = n! / (r!(n-r)!)

In this case, n = 11 (the total number of tax returns) and r = 3 (the number of tax returns to be selected for the audit).

Using the combination formula, the calculation would be:

11C3 = 11! / (3!(11-3)!)
= 11! / (3!8!)
= (11*10*9) / (3*2*1)
= 165

Therefore, there are 165 ways for the IRS auditor to select 3 of the 11 tax returns for an audit.

For the second question, to find the number of different subcommittees that can be formed from a board of directors with 8 members choosing 3 members, you can use the combination formula again.

nCr = n! / (r!(n-r)!)

In this case, n = 8 (the total number of board members) and r = 3 (the number of members needed for the subcommittee).

Using the combination formula:

8C3 = 8! / (3!(8-3)!)
= 8! / (3!5!)
= (8*7*6) / (3*2*1)
= 56

Therefore, there are 56 different subcommittees possible.

For the third question, to find the number of license plates that can be made using 3 letters followed by 3 digits with repetition allowed, you need to determine the number of choices for each position.

For the first letter, there are 26 options (A-Z).
For the second letter, there are also 26 options.
For the third letter, again, there are 26 options.

For the first digit, there are 10 options (0-9).
For the second digit, 10 options.
For the third digit, 10 options.

To find the total number of plates, multiply the number of possibilities for each position:

26 * 26 * 26 * 10 * 10 * 10 = 17,576,000

Therefore, there are 17,576,000 possible license plates that can be made.

For the fourth question, to find the probability of selecting 3 members from town A and 2 members from town B out of a randomly selected council of 5 members, you need to calculate the total number of possible council combinations and the number of combinations with 3 members from town A and 2 members from town B.

Total number of possible council combinations:
The total number of people to choose from is 8 (town A members) + 7 (town B members) + 5 (town C members) = 20 people.
Using the combination formula:

20C5 = 20! / (5!(20-5)!)
= 20! / (5!15!)
= (20*19*18*17*16) / (5*4*3*2*1)
= 15,504

Number of combinations with 3 members from town A and 2 members from town B:
To select 3 members from town A, you have 8C3 = 8! / (3!(8-3)!) = 56 combinations.
To select 2 members from town B, you have 7C2 = 7! / (2!(7-2)!) = 21 combinations.
Multiply these two combinations together to get the total number of combinations with 3 members from town A and 2 members from town B:
56 * 21 = 1,176

Probability = Number of desired combinations / Total number of possible combinations
Probability = 1176 / 15504
Probability ≈ 0.0757 (rounded to four decimal places)

Therefore, the probability that 3 members are from town A and 2 members are from town B is approximately 0.0757.

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