Problem 31 - A Graphical Approach to Precalculus with Limits

Question

Use a calculator to evaluate each expression. $$ { }_{15} C_{8} $$

Step-by-Step Solution

Answer

The value of \( {}_{15}C_8 \) is 6435.

Step 1: Understand the Problem

The problem is asking us to evaluate the combination \( {}_{15}C_8 \) which represents the number of ways to choose 8 items from a set of 15 without regard to order.

Step 2: Apply the Combination Formula

The formula for a combination is given by \( {}_{n}C_{r} = \frac{n!}{r!(n-r)!} \). In our case, \( n = 15 \) and \( r = 8 \). Substitute these values into the formula: \( {}_{15}C_8 = \frac{15!}{8!\cdot(15-8)!} \).

Step 3: Calculate Factorials

Calculate the necessary factorials: \( 15! = 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8! \) and \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \). Notice that \( 8! \) will cancel in the computation.

Step 4: Simplify the Expression

Plug into the equation: \( {}_{15}C_8 = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7!} \). Simplify by canceling out the \( 8! \) and compute the remaining part.

Step 5: Use the Calculator

Divide the result of the numerator by \( 7! = 5040 \). Use a calculator to get the final result: \( \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{5040} = 6435 \).

Step 6: Round to Final Answer

Double-check the calculation using a calculator to ensure accuracy. The value of \( {}_{15}C_8 \) is confirmed to be 6435.

Key Concepts

Factorials

Factorials are a crucial concept when dealing with combinations and permutations. In simple terms, the factorial of a non-negative integer \( n \), denoted \( n! \), is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials grow very fast with increasing \( n \) and are fundamental in calculating combinations and permutations.

They are used because they help in counting arrangements and selections, ensuring every possible order or selection is counted correctly. In the given problem, we compute several factorials, \( 15! \) and \( 7! \), which act as building blocks in the combination formula.

If you use a calculator, most scientific calculators have a factorial button, often marked as \( ! \). This helps to avoid manually computing large factorials, which can be time-consuming and prone to error.

Combination Formula

The combination formula is the mathematical expression used to find the number of ways to choose \( r \) items from \( n \) items without considering the order of selection. It is represented as \( {}_{n}C_{r} \), or sometimes as \( C(n, r) \). The formula is given by:

\[ {}_{n}C_{r} = \frac{n!}{r!(n-r)!} \]

This formula calculates combinations by dividing the total number of permutations by the number of permutations for each selected subgroup \( r! \) and the remaining items \( (n-r)! \).

In the exercise, we evaluate \( {}_{15}C_{8} \), meaning choosing 8 items from a group of 15. By substituting 15 for \( n \) and 8 for \( r \) into the formula, we find the number of combinations possible. This involves calculating \( 15! \), dividing by both \( 8! \) and \( 7! \), simplifying the expression, and computing the result to get the answer of 6435 combinations.

Permutations

Permutations refer to different arrangements of a set of items where the order does matter, unlike combinations where the order doesn't matter. The formula for permutations is slightly different because permutations consider each potential order as unique.

For \( n \) total items and choosing \( r \) of them, the number of permutations is given by:

\[ P(n, r) = \frac{n!}{(n-r)!} \]

Note that \( r! \) is not in the denominator like in combinations because every arrangement is counted individually. For example, in arranging 3 books out of 5 on a shelf, each order leads to a different permutation. Hence, \( P(5, 3) = \frac{5!}{2!} \).

While this specific exercise does not require calculating permutations, understanding the difference between permutations and combinations provides deeper insight into how selections and arrangements are computed in mathematics. Both concepts use factorials, but in different ways depending on whether order matters or not.