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.

Please submit one question per post.

The first answer is 11*10*9/(1*2*3)= 165,
since the order they are picked does not matter.

To answer these questions, we can use the concept of combinations, permutations, and probability. Let's break down each question and explain how to solve them.

1. How many ways can an IRS auditor select 3 of 11 tax returns for an audit?
To calculate this, we can use the formula for combinations. The formula is:
C(n, r) = n! / (r!(n-r)!)
Here, n is the total number of items (11 tax returns) and r is the number of items being selected (3 tax returns).

So, applying the formula:
C(11, 3) = 11! / (3!(11-3)!) = 11! / (3!8!) = (11 * 10 * 9) / (3 * 2 * 1) = 165

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

2. There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?
In this case, we need to find the number of combinations again.

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

Hence, there are 56 different subcommittees possible.

3. 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?
Since repetition is allowed, we have multiple choices for each character position.

For the letters, there are 26 choices for each position, so there are 26^3 = 26 * 26 * 26 = 17,576 ways to choose the letters.

For the digits, there are 10 choices for each position, so there are 10^3 = 10 * 10 * 10 = 1,000 ways to choose the digits.

To get the total number of license plates, we multiply the two possibilities together:
17,576 * 1,000 = 17,576,000

Therefore, there can be 17,576,000 different license plates made with repetition of letters and digits allowed.

4. 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.
To find the probability, we can use the concept of combinations.

First, we need to determine the number of ways to select 3 people from town A and 2 people from town B.

Using the formula for combinations with repetition:
C(8, 3) * C(7, 2) = (8! / (3!(8-3)!)) * (7! / (2!(7-2)!)) = (56 * 21) = 1,176

We then need to find the total number of ways to select any 5 people from the entire group of 20 individuals:

C(20, 5) = 20! / (5!(20-5)!) = 20! / (5!15!) = 15,504

The probability is calculated by:

Probability = (Number of favorable outcomes) / (Number of total possible outcomes)
= 1,176 / 15,504
= 0.0758 (rounded to four decimal places)

So, the probability that 3 people are from town A and 2 are from town B is approximately 0.0758.