Problem 55 - Algebra for College Students

Question

How many different juries consisting of 11 people can be chosen from a group of 30 people? \(\quad 54,627,300\)

Step-by-Step Solution

Answer

There are 54,627,300 ways to choose 11 jurors from 30 people.

Step 1: Understanding Combinations

To solve this problem, we need to understand that selecting juries consists of finding the number of combinations (not permutations) because the order of selection does not matter. This is a classic combination problem where we select a subset (jury) from a larger set (group of people).

Step 2: Using the Combination Formula

The formula for computing combinations, also known as binomial coefficients, is given by: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]where \(n\) is the total number of items to choose from, \(r\) is the number of items to choose, and \(!\) denotes factorial, which is the product of all positive integers up to that number.

Step 3: Substituting Values into the Formula

In our problem, we have \(n = 30\) and \(r = 11\). Substitute these values into the formula:\[ C(30, 11) = \frac{30!}{11!(30-11)!} = \frac{30!}{11! \times 19!} \]

Step 4: Calculating Factorials

Now we need to calculate the factorials:- \(30!\) is 30 times 29 times 28 and so on down to 1.- \(11!\) is 11 times 10 times 9 and so on down to 1.- \(19!\) is 19 times 18 times 17 and so on down to 1.

Step 5: Simplifying the Expression

To simplify the factorial calculations, we notice:\[\frac{30!}{19!} = 30 \times 29 \times 28 \times \ldots \times 20 \]This cancels out the \(19!\) factor in the denominator.

Step 6: Final Calculation

Now compute the simplified expression:\[ C(30, 11) = \frac{30 \times 29 \times 28 \times 27 \times 26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \] Calculating this gives:\[ C(30, 11) = 54,627,300 \]

Step 7: Conclusion

Thus, the number of different juries consisting of 11 people that can be chosen from a group of 30 people is 54,627,300.

Key Concepts

Factorials

The concept of factorials is crucial when working with combinations and permutations. A factorial, denoted by the symbol \(!\), is the product of all positive integers less than or equal to a given number. For any positive integer \(n\), its factorial is defined as:

• \( n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \)

For example:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)

Factorials grow very quickly with the increase of numbers. However, in problems like jury selection, calculations are simplified by employing factorials in the combination formula, allowing cancelation of terms for easier computation.

Binomial Coefficients

Binomial coefficients are a central part of combinatorics, often represented by \( C(n, r) \) or \( \binom{n}{r} \). They are used to count the number of ways to choose \( r \) subsets from a set containing \( n \) elements, without regard to the order of selection.The binomial coefficient is defined as:\[ C(n, r) = \frac{n!}{r!\,(n-r)!} \]This formula calculates how many ways you can choose \( r \) items, where \( ! \) denotes a factorial. Given the order does not matter in these selections, this formula helps us efficiently solve combination-related problems, like forming juries, teams, or groups from larger pools.

Combination Formula

The combination formula plays a vital role when solving problems that involve selecting a subset from a larger set without attention to the sequence. As given in the jury selection problem, we use the formula:\[ C(n, r) = \frac{n!}{r!(n-r)!} \]When solving a problem with this formula:

• \( n \) represents the total number of people or items available.
• \( r \) is the number we wish to select.
• Factorials help break down the large multiplications into manageable calculations by canceling out common terms.

For example, selecting 11 jurors from 30 can be calculated by substituting \( n = 30 \) and \( r = 11 \) into the formula, simplifying through factorial reduction and division.

Jury Selection Problem

The jury selection problem is a practical scenario that highlights the application of combinations and binomial coefficients. It asks us to determine the number of possible groups, or juries, that can be formed out of a larger population. This type of problem is common in various fields beyond jury selection, such as optimization and decision-making processes. In this specific exercise, when choosing 11 jurors from 30 individuals, we are interested in how many different ways this can be done. Since the order does not matter (it doesn't matter if John is chosen before Jane), it's a combination problem rather than permutation. Hence, using binomial coefficients and the combination formula, we found that there are 54,627,300 unique ways to form an 11-member jury from a pool of 30 candidates, illustrating the abundance of choices in selection processes without regard to order. This understanding is critical when making decisions that require sorted, non-ordered selections.