Problem 47 - College Algebra with Corequisite Support

Question

How many ways can a baseball coach arrange the order of 9 batters if there are 15 players on the team?

Step-by-Step Solution

Answer

There are 1,814,400,000 ways to arrange the 9 batters from 15 players.

Step 1: Understanding the Problem

We need to determine in how many different ways a coach can arrange 9 batters out of 15 players.

Step 2: Choosing 9 Players from 15

The first step is to calculate how many ways we can choose 9 players from a total of 15. This can be done using combinations because the order in which we select the players does not matter yet. The formula for combinations is: \( C(n, k) = \frac{n!}{k!(n-k)!} \). Here, \( n = 15 \) and \( k = 9 \).

Step 3: Calculating Combinations

Using the formula: \[ C(15, 9) = \frac{15!}{9! \cdot (15 - 9)!} \]. We need to calculate the value as: \( \frac{15!}{9! \cdot 6!} \).

Step 4: Simplifying Factorials

Calculate the factorial values needed: \( 15! = 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9! \); \( 9! = 9 \times 8 \times ... \times 1 \); and \( 6! = 6 \times 5 \times ... \times 1 \). Simplify the expression \( \frac{15!}{9! \cdot 6!} \) by cancelling out \( 9! \).

Step 5: Finding the Number of Arrangements

After choosing 9 players, we need to arrange them in order. This is a permutation of 9 players, calculated as \( 9! \). The total number of ways is the product of the combination from Step 3 and this permutation: \[ C(15, 9) \times 9! \].

Step 6: Calculating Final Result

Compute \( C(15, 9) = 5005 \) and then multiply by \( 9! = 362880 \). Thus, the total arrangements are \( 5005 \times 362880 = 1,814,400,000 \).

Key Concepts

Permutations

A permutation is an arrangement of objects, where the order does matter. When arranging items or placing them in specific sequences, we're dealing with permutations. The key question is how many different sequences or orders can be formed with a given set of items.

To calculate permutations, we use the factorial concept. If you have `n` distinct objects and want to find how many ways you can order all of them, you'll calculate `n!`. For example, if there are 9 players, the permutations of these players arranged in a specific order would be denoted as `9!`.

• Permutations are used when the order is important.
• The formula for permutations of `n` items taken `k` at a time is represented as:
  \[ P(n, k) = \frac{n!}{(n-k)!} \]
• In our case, choosing and arranging the 9 batters involves calculating just `9!` since we're arranging all chosen players.

Combinations

Combinations are selections of items where the order does not matter. For example, if you are forming a group from a pool of players, it does not matter in what order they are chosen, only which players make it into the group. This is where combinations differ significantly from permutations.

The formula to calculate combinations is:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
Here, `n` is the total number of items to choose from, and `k` is the number of items to select. Using combinations is ideal when you're interested in the set formed rather than the sequence.

• Order does not matter in combinations.
• Combinations involve division to eliminate redundancies from order.
• In our exercise, we calculated how to choose 9 players from 15 using \( C(15, 9) \).

Factorials

Factorials are a mathematical concept that plays a pivotal role in permutations and combinations. A factorial of a non-negative integer `n`, denoted by `n!`, is the product of all positive integers less than or equal to `n`. Factorials are defined for non-negative integers only, and they provide the foundation for calculations involving permutations and combinations.

For instance, the factorial of 3 is:

• \( 3! = 3 \times 2 \times 1 = 6 \)

Factorials grow rapidly with larger `n`, which is why things like simplifying and cancelling are essential in combination and permutation computations.

• Factorials are key for setting up permutation (\( n! \)) and combination (\( \frac{n!}{k!(n-k)!} \)) calculations.
• Simplification often involves cancelling factorial values, such as \( \frac{15!}{9! \cdot 6!} \) in our exercise example.