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In the U.S. Senate, there are 21 members on the Committee on Banking, Housing, and Urban Affairs. Nine of these 21 members are selected to be on the Subcommittee on Economic Policy. How many different committee structures are possible for this subcommittee?

Short Answer

Expert verified
6,604,203

Step by step solution

01

- Understand the problem

Determine the total number of ways to choose 9 members out of 21 members.
02

- Use Combinatorics Formula

To choose 9 members out of 21, use the combination formula which is: \[ C(n, k) = \frac{n!}{k! (n - k)!} \]
03

- Plug in the values

In our problem, n = 21 and k = 9: \[ C(21, 9) = \frac{21!}{9! (21 - 9)!} = \frac{21!}{9! \cdot 12!} \]
04

- Simplify the factorials

Simplify the factorial expression: \[ C(21, 9) = \frac{21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12!}{9! \times 12!} \]
05

- Cancel common terms

Cancel the common \(12!\) terms in the numerator and denominator: \[ C(21, 9) = \frac{21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13}{9!} \]
06

- Calculate the numerical value

Calculate the numerical value by first evaluating the numerator and then dividing by \(9!\): Numerator: \( 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 = 2,432,902,008,176,640 \)\( and Denominator: \( 9! = 362,880 \)\) So, \[ C(21, 9) = \frac{2,432,902,008,176,640}{362,880} = 6,604,203 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

combinations
Combinations are a way to select items from a larger set where the order does not matter. This is different from permutations where the order does matter. To calculate the number of combinations, we use a mathematical formula. The combination formula is written as:

\( C(n, k) = \frac{n!}{k!(n - k)!} \)

Here, \( n \) is the total number of items, and \( k \) is the number of items we want to choose. This formula counts all possible ways to pick \( k \) items from \( n \) items without caring about the order.

For example, if we have 5 fruits and want to pick 2, we would write:

\( C(5, 2) = \frac{5!}{2!(5 - 2)!} = \frac{5!}{2! \cdot 3!} = 10 \)

This says there are 10 different ways to pick 2 fruits out of 5.
factorial
The factorial function, denoted by an exclamation point (!), is a way to calculate the product of all positive integers up to a specified number. If you see \( n! \), that means:

\( n! = n \cdot (n - 1) \cdot (n - 2) \cdot \ldots \cdot 1 \)

For example, 4! is:

\( 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \)

Factorials show up often in combinatorics when calculating combinations or permutations. They help count the number of ways items can be arranged or selected. Factorials grow very quickly. For example, 10! is 3,628,800, which is a large number.
subcommittee selection
In many real-life situations, we need to form smaller groups or subcommittees from a larger pool. One example is choosing a subcommittee from a committee, just like in our problem with the U.S. Senate committee.

When forming subcommittees, combinations are useful because the order in which the subcommittee members are chosen doesn't matter. What counts is only who is in the group.

Suppose we want to pick 9 members from a committee of 21 members. The number of ways to do this is calculated using combinations:

\( C(21, 9) = \frac{21!}{9!(21 - 9)!} = \frac{21!}{9! \cdot 12!} \)

By solving this calculation, we find the total number of different ways to form this 9-member subcommittee.
binomial coefficient
The binomial coefficient is another name for combinations and is denoted as \( \binom{n}{k} \). It describes the number of ways to choose \( k \) items from \( n \) items, regardless of order. The binomial coefficient uses the same formula as combinations:

\( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

This concept is widely used in areas like probability, statistics, and algebra. For instance, in the binomial theorem, which expands expressions like \( (a + b)^n \), the coefficients are given by the binomial coefficients.

If we want to find how many ways we can choose 3 objects out of 7, we write:

\( \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!} \)

Calculating this gives us the number of ways to choose 3 items from 7.

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Most popular questions from this chapter

Suppose that a shipment of 120 electronic components contains 4 defective components. To determine whether the shipment should be accepted, a quality- control engineer randomly selects 4 of the components and tests them. If 1 or more of the components is defective, the shipment is rejected. What is the probability that the shipment is rejected?

Suppose that a poll is being conducted in the village of Lemont. The pollster identifies her target population as all residents of Lemont 18 years old or older. This population has 6494 people. (a) Compute the probability that the first resident selected to participate in the poll is Roger Cummings and the second is Rick Whittingham. (b) The probability that any particular resident of Lemont is the first person picked is \(\frac{1}{6494} .\) Compute the probability that Roger is selected first and Rick is selected second, assuming independence. Compare your results to part (a). Conclude that, when small samples are taken from large populations without replacement, the assumption of independence does not significantly affect the probability.

A family has six children. If this family has exactly two boys, how many different birth and gender orders are possible?

In finance, a derivative is a financial asset whose value is determined (derived) from a bundle of various assets, such as mortgages. Suppose a randomly selected mortgage has a probability of 0.01 of default. (a) What is the probability a randomly selected mortgage will not default (that is, pay off)? (b) What is the probability a bundle of five randomly selected mortgages will not default assuming the likelihood any one mortgage being paid off is independent of the others? Note: A derivative might be an investment in which all five mortgages do not default. (c) What is the probability the derivative becomes worthless? That is, at least one of the mortgages defaults? (d) In part (b), we made the assumption that the likelihood of default is independent. Do you believe this is a reasonable assumption? Explain.

Suppose that a satellite defense system is established in which four satellites acting independently have a 0.9 probability of detecting an incoming ballistic missile. What is the probability that at least one of the four satellites detects an incoming ballistic missile? Would you feel safe with such a system?

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