Problem 7 - Algebra 2

Question

Determine whether each situation involves a permutation or a combination. Then find the number of possibilities. an arrangement of 4 blue tiles, 2 red tiles, and 3 black tiles in a row

Step-by-Step Solution

Answer

This situation involves a permutation with 1260 possibilities.

Step 1: Understanding the Situation

The problem asks us to determine if this is a permutation or combination problem. An 'arrangement' of items generally implies that the order matters, thus indicating a permutation.

Step 2: Defining the Permutation

Since this is a permutation problem where we have repeated items, we need to use the formula for permutations of multiset. For an arrangement of a group of items with repeats, the formula is: \[ \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdot \ldots} \] where \( n \) is the total number of items, and \( n_1, n_2, n_3, \ldots \) are the counts of each repeated item.

Step 3: Identifying the Values

Here, the total number of tiles is \( n = 4 + 2 + 3 = 9 \). The number of blue tiles is 4, the number of red tiles is 2, and the number of black tiles is 3. These will be used as \( n_1, n_2, \) and \( n_3 \) respectively.

Step 4: Applying the Formula

Plug the values into the permutation formula: \[ \frac{9!}{4! \cdot 2! \cdot 3!} \]

Step 5: Calculating Factorials

Calculate the factorial for each number. - \( 9! = 362880 \)- \( 4! = 24 \)- \( 2! = 2 \)- \( 3! = 6 \)

Step 6: Performing the Calculation

Substitute the factorial values into the formula: \[ \frac{362880}{24 \cdot 2 \cdot 6} = \frac{362880}{288} \] Calculate the final result: \( 1260 \).

Step 7: Concluding the Solution

There are 1260 different ways to arrange the tiles.

Key Concepts

Multiset Permutations

When we talk about multiset permutations, we refer to arranging a large group of items where some may repeat. This differs from regular permutations where all items are distinct. Here, the repetition changes the way we calculate possibilities.
For example, in the exercise, the total number of tiles is 9, with varying quantities of each color. Instead of calculating as if each tile were unique, we use a special formula for multiset permutations:

• The total number of items is defined as \( n \).
• Each set of identical items is labeled \( n_1, n_2, \) etc.

Thus, the formula is:\[\frac{n!}{n_1! \cdot n_2! \cdot \ldots}\]This formula adjusts the total number of arrangements to account for repeated items, ensuring that swapped identical items aren't counted multiple times. For our set of tiles, we calculated as follows: use \( n = 9 \) (total tiles) and divide it by the factorials of each type of tile: \[ \frac{9!}{4! \cdot 2! \cdot 3!} \] to find the different possible arrangements.

Factorials

Factorials are a core mathematical concept used in permutations and combinations. A factorial, denoted as \( n! \), is the product of all positive integers up to \( n \).
For example, \( 4! \) means multiplying all numbers from 1 to 4:

• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 2! = 2 \times 1 = 2 \)

Factorials play a crucial role in calculating permutations and combinations since they provide the means to calculate how many ways items can be arranged in order.
In the exercise, we used factorials to determine how the tiles could be ordered differently and solved:\[ \frac{362880}{24 \times 2 \times 6} = 1260\] This calculation indicated the total number of different sequences possible.

Combinations vs Permutations

Understanding the difference between combinations and permutations is key in determining how to approach a problem involving arrangement or selection.

• Permutations are about arranging items where the order matters. For example, arranging alphabets in different ways to form words.
• Combinations, on the other hand, are about selecting items where the order doesn't matter. An example is selecting a committee from a group of people.

In permutations, if we have 3 distinct items \( A, B, \) and \( C \), they can be arranged in multiple orders: ABC, ACB, BAC, BCA, CAB, CBA. Combinations would simply refer to the selection of any two, like \( AB, BC, \) or \( AC \), without regard to arrangement.
In the tile exercise, we considered permutations because the arrangement affects the overall outcome. Each different sequence of tiles counts as a different possibility, justifying our use of the multiset permutation formula.