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A committee of 7, consisting of 2 Republicans, 2 Democrats, and 3 Independents, is to be chosen from a group of 5 Republicans, 6 Democrats, and 4 Independents. How many committees are possible?

Short Answer

Expert verified
There are 600 possible committees that can be formed from the given number of Republicans, Democrats, and Independents. This can be calculated using combinations, by finding the number of ways to choose 2 Republicans, 2 Democrats, and 3 Independents, and then multiplying the results together: \( 10 \times 15 \times 4 = 600 \).

Step by step solution

01

Find the number of combinations for Republicans

We need to choose 2 Republicans from a group of 5 Republicans. The formula for combinations is: \[ C(n, k) = \frac{n!}{k!(n - k)!} \] Where "n" represents the total number of items in the group, "k" represents the number of items we want to choose, and "!" denotes the factorial. In this case, we will have: n = 5 (number of Republicans) and k = 2 (number of Republicans to choose). Applying the formula, we get: \[ C(5, 2) = \frac{5!}{2!(5-2)!} \\ = \frac{5!}{2!3!} \\ = \frac{120}{(2)(6)} \\ = 10 \] So there are 10 possible ways to choose 2 Republicans.
02

Find the number of combinations for Democrats

We need to choose 2 Democrats from a group of 6 Democrats. Here n = 6 (number of Democrats) and k = 2 (number of Democrats to choose). Applying the formula, we get: \[ C(6, 2) = \frac{6!}{2!(6-2)!} \\ = \frac{6!}{2!4!} \\ = \frac{720}{(2)(24)} \\ = 15 \] So there are 15 possible ways to choose 2 Democrats.
03

Find the number of combinations for Independents

We need to choose 3 Independents from a group of 4 Independents. Here n = 4 (number of Independents) and k = 3 (number of Independents to choose). Applying the formula, we get: \[ C(4, 3) = \frac{4!}{3!(4-3)!} \\ = \frac{4!}{3!1!} \\ = \frac{24}{(6)(1)} \\ = 4 \] So there are 4 possible ways to choose 3 Independents.
04

Calculate total number of possible committees

Now that we know the number of possible ways to choose Republicans, Democrats, and Independents, we can calculate the total number of possible committees by multiplying the results together: Total committees = (Number of combinations for Republicans) × (Number of combinations for Democrats) × (Number of combinations for Independents) Total committees = 10 × 15 × 4 = 600 There are 600 possible committees that can be formed from the given number of Republicans, Democrats, and Independents.

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Key Concepts

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

Factorial Notation
Understanding factorial notation is essential in solving many probability and combinatorics problems. The factorial of a non-negative integer, denoted by an exclamation point (!), is the product of all positive integers up to that number. For example, the factorial of 5, written as (5!), is calculated as (5! = 5 × 4 × 3 × 2 × 1 = 120).

Factorial notation simplifies the expression of large products and is especially useful when calculating combinations and permutations, where ordering or groupings of items are considered. Zero factorial is defined to be (0! = 1).

The factorial function grows rapidly with larger numbers, so when dealing with combinations in probability problems, factorials help express calculations compactly. In the given exercise, factorials were used to calculate how many ways committees can be formed, exemplifying the practical use of factorial notation in determining the number of possible arrangements.
Combinatorial Analysis
Combinatorial analysis is a field of mathematics concentrated on counting, arrangement, and combination of elements in sets, without considering the order. It is widely applied in fields such as probability theory, statistics, and algorithm design.
  • Combinations: When the order doesn't matter and repetition is not allowed.
  • Permutations: When the order does matter.


This analytical approach allows us to solve problems where we are interested in how many ways a certain number of items can be selected from a larger set. An example of combinatorial analysis in the exercise was determining the number of possible committees by finding the number of combinations of Republicans, Democrats, and Independents that could be chosen.

By applying combinatorial formulas, the complexity of counting is significantly reduced, leading to quicker and more accurate solutions in probabilistic and combinatorial problems.
Permutations and Combinations
Permutations and combinations are fundamental concepts in the study of combinatorial analysis and probability. While they might seem similar, they serve different purposes:
  • Permutations: This refers to the arrangement of all or part of a set of items where the order is important. For example, the permutation of three different books on a shelf can be arranged in different sequences, each sequence being a unique permutation.
  • Combinations: Unlike permutations, combinations refer to selecting items from a set where the order does not matter. For instance, when forming a committee, it doesn’t matter which order members are chosen, only the group as a whole.


The exercise given illustrates the use of combinations as the specific arrangement of members in the committee is irrelevant, only the composition of the group matters. Understanding when to use permutations or combinations is crucial in solving problems related to probability and is often indicated by the context of the problem itself whether the order of selection is a significant factor.

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Most popular questions from this chapter

A dance class consists of 22 students, 10 women and 12 men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible?

Consider a tournament of \(n\) contestants in which the outcome is an ordering of these contestants, with ties allowed. That is, the outcome partitions the players into groups, with the first group consisting of the players that tied for first place, the next group being those that tied for the next best position, and so on. Let \(N(n)\) denote the number of different possible outcomes. For instance, \(N(2)=3\) since in a tournament with 2 contestants, player 1 could be uniquely first, player 2 could be uniquely first, or they could tie for first. (a) List all the possible outcomes when \(n=3\). (b) With \(N(0)\) defined to equal 1 , argue, without any computations, that $$ N(n)=\sum_{i=1}^{n}\left(\begin{array}{l} n \\ i \end{array}\right) N(n-i) $$ HINT: How many outcomes are there in which \(i\) players tie for last place? (c) Show that the formula of part (b) is equivalent to the following: $$ N(n)=\sum_{i=0}^{n-1}\left(\begin{array}{l} n \\ i \end{array}\right) N(i) $$ (d) Use the recursion to find \(N(3)\) and \(N(4)\).

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