Problem 4 - Algebra 2

Question

Evaluate each expression. \(C(6,1)\)

Step-by-Step Solution

Answer

There are 6 ways to choose 1 object from a set of 6 objects.

Step 1: Understanding the Combination Notation

The expression \(C(n, r)\) represents the number of combinations, or ways to choose \(r\) objects from a set of \(n\) objects without regard to order. Here, \(C(6, 1)\) indicates that we want to choose 1 object from a set of 6 objects.

Step 2: Applying the Combination Formula

The formula for calculating combinations is \(C(n, r) = \frac{n!}{r!(n-r)!}\). In our problem, \(n = 6\) and \(r = 1\). Substituting these values into the formula gives us: \(C(6, 1) = \frac{6!}{1!(6-1)!} = \frac{6!}{1! \times 5!}\).

Step 3: Simplifying Factorials

To simplify \(\frac{6!}{1! \times 5!}\), recall that a factorial \(n!\) represents the product of all positive integers up to \(n\). So, \(6! = 6 \times 5!\). Cancel out the \(5!\) in the numerator and the denominator, leaving: \(\frac{6 \times 5!}{1! \times 5!} = \frac{6}{1} = 6\).

Step 4: Concluding the Calculation

After simplifying, we find that \(C(6, 1) = 6\). This means there are 6 ways to choose 1 object from a set of 6 objects.

Key Concepts

Factorials

A factorial, represented by the symbol \(!\), is a mathematical operation that multiplies a number by every positive integer lesser than itself. For example, to calculate \(n!\), you multiply \(n\) by \(n-1\), then by \(n-2\), continuing this process until you multiply by 1. The product of these numbers gives you the factorial of \(n\).

This can be mathematically illustrated as:

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

Factorials play a crucial role in many mathematical concepts, especially in permutations and combinations, because they help determine the number of ways in which objects or numbers can be arranged. It's also important to remember that \(0!\) is defined to be 1 by convention. This is essential in maintaining consistent mathematical rules while solving problems involving combinations.

Combination Formula

The combination formula helps us calculate how many ways we can choose a subset of items from a larger set, without regard to the order of selection. This is expressed as \(C(n, r)\) or sometimes \(nCr\), where \(n\) is the total number of items, and \(r\) is the number of items to choose.

The formula is given by:
\[C(n, r) = \frac{n!}{r!(n-r)!}\]

Here, the factorial operation from our previous section assists in breaking down this equation:

• In the numerator, \(n!\) signifies all the ways to arrange all \(n\) items.
• In the denominator, \(r!\) accounts for the repetition within the selected subset.
• The component \((n - r)!\) accounts for the items that are not chosen in the subset.

Remember, combinations differ from permutations because the order of items does not matter in combinations, making it a crucial piece for solving problems like our original exercise.

Permutations

In permutations, the order of selection does matter. If you think of permutations, imagine arranging a set of books on a shelf, where each arrangement counts differently if the order is changed.

Mathematically, permutations of selecting \(r\) items from \(n\) items is expressed as \(P(n, r)\) and is calculated by:
\[P(n, r) = \frac{n!}{(n-r)!}\]

The formula shows that:

• The numerator \(n!\) again represents all possible arrangements of \(n\) items.
• The denominator \((n-r)!\) eliminates the arrangements of items outside the selected \(r\), highlighting the importance of order.

The key distinction here is the significance of arrangement. In our original problem, we dealt with combinations where order does not matter, unlike permutations. Understanding both concepts can greatly enhance your problem-solving toolkit in various fields like statistics, computer science, and probability.