Problem 33 - Algebra and Trigonometry

Question

33–40 These problems involve distinguishable permutations. Arrangements In how many ways can two blue marbles and four red marbles be arranged in a row?

Step-by-Step Solution

Answer

There are 15 ways to arrange the two blue marbles and four red marbles in a row.

Step 1: Understanding the Problem

We have two blue marbles and four red marbles, and we want to find out how many distinct ways we can arrange these six marbles in a row. Each blue marble is identical to the other blue marble, and similarly, each red marble is identical to the other red marbles.

Step 2: Total Arrangement Calculation

If all six marbles were different, the total number of arrangements would simply be the factorial of the total number of marbles. Thus, we calculate it as follows: \[6! = 720\]

Step 3: Account for Indistinguishable Marbles

Since the marbles of the same color are indistinguishable, we must divide by the factorials of the number of indistinguishable marbles. Thus for the blue marbles: \[2! = 2\] and for the red marbles:\[4! = 24\]

Step 4: Final Calculation for Distinguishable Arrangements

Divide the total number of arrangements by the product of the factorials of the indistinguishable items to get the number of unique arrangements:\[\frac{6!}{2! \times 4!} = \frac{720}{2 \times 24} = \frac{720}{48} = 15\]

Step 5: Verify the Final Answer

Verify that the numbers seem consistent: total permutations of distinct items was 720, and dividing by the 48 ways caused by redundancies from indistinguishable marbles seems logical, leading to an integer solution.

Key Concepts

Factorial Calculations

Factorial calculations are a fundamental part of finding permutations and combinations. A factorial is represented as \( n! \) and means that you multiply all positive integers from 1 to \( n \). For example, \( 6! \) is calculated as \( 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \). This tells us the total number of ways we can arrange 6 different objects in a sequence.

Factorials are important because they help describe how many possible arrangements can be made. Whenever you're handling problems involving arranging objects, factorial calculations lay the foundation of understanding the possible permutations.

• Always determine the total number of objects you want to arrange.
• Use factorials to calculate all possible sequences if the objects were distinguishable.

Indistinguishable Objects

When dealing with arrangements, sometimes objects are indistinguishable or identical within a set. This means you cannot tell them apart by looking at them, like identical twin marbles. In these cases, we must adjust our calculations to avoid overcounting the arrangements.

To account for indistinguishable objects in arrangements, you divide the total permutations by the factorial of each set of identical objects. For example, if you have 2 identical blue marbles and 4 identical red marbles, you would calculate:

• Blue marbles: \( 2! = 2 \)
• Red marbles: \( 4! = 24 \)

This step is crucial because it adjusts the total arrangements down to reflect only unique sequences of arrangement. When objects are indistinguishable, they don't contribute to changing the sequence's unique character as distinct objects would.

Remember, when objects are indistinguishable, they need special consideration in permutation problems to avoid counting the same sequence more than once.

Permutation Formula

The permutation formula is essential when it comes to finding the number of ways to arrange objects in a sequence, specifically when some of these objects are indistinguishable. The general formula for permutations of \( n \)objects where there are groups of indistinguishable objects is: \[ \frac{n!}{n_1! \times n_2! \times ... \times n_k!} \] here, \( n! \) is the factorial of the total number of objects, and each \( n_i! \) represents the factorial calculation for each set of indistinguishable objects.

For example, in the problem involving 6 marbles, we wanted to know how many ways to arrange them if there were two blue and four red marbles. Using the permutation formula, the problem solves as follows:

• Total marbles factorial: \( 6! = 720 \)
• Blue marbles: \( 2! = 2 \)
• Red marbles: \( 4! = 24 \)

Plugging into the formula, we find the number of distinct arrangements as \[ \frac{6!}{2! \times 4!} = \frac{720}{48} = 15 \].

This formula gives a proper count of sequences considering the indistinguishable nature of some items, ensuring that each arrangement is truly unique. Whenever you who deal with distinguishable and indistinguishable objects in permutation problems, remember the power ofthis formula.