Problem 78 - Algebra and Trigonometry Real Mathematics, Real People

Question

Evaluate \(_{n} C_{r} .\) Verify your result using a graphing utility. $$_{9} C_{5}$$

Step-by-Step Solution

Answer

The value of \(_{9} C_{5}\) is 126.

Step 1: Identify the variables in the combination formula

In the combination formula \(_{n} C_{r} = \frac{n!}{r!(n-r)!}\), we can identify that in our exercise \(n = 9\) and \(r = 5\).

Step 2: Substitute the variables into the combination formula

Substituting \(n = 9\) and \(r = 5\), the formula becomes: \(_{9} C_{5} = \frac{9!}{5!(9-5)!}.\)

Step 3: Simplify factorials and subtract in the denominator

We simplify the expression to: \(_{9} C_{5} = \frac{9!}{5!(4)!}\). Next, we calculate the factorial expressions: \(9! = 362880\), \(5! = 120\), and \(4! = 24\).

Step 4: Complete the calculations

By substituting back we get: \(_{9} C_{5} = \frac{362880}{120 * 24} = 126.\

Key Concepts

Factorials

Understanding factorials is crucial in various branches of mathematics, particularly when dealing with combinations and permutations. A factorial, denoted by an exclamation mark (!), is the product of all positive integers up to a given number. For example, the factorial of 5, written as \(5!\), is calculated by multiplying \(5 \times 4 \times 3 \times 2 \times 1 = 120\).

When working with factorials in algebraic expressions, it is often necessary to simplify the factorials by canceling out common factors. For instance, in the combination \(_{9}C_{5}\), we simplify \(9!\) as \(9 \times 8 \times 7 \times 6 \times 5!\), allowing us to cancel the \(5!\) in the numerator and denominator. Understanding this technique can vastly simplify calculations in combinatorics.

Binomial Coefficients

The binomial coefficient, often read as 'n choose r', represents the number of ways to choose r elements from a set of n elements without regard to order. In algebra, it is denoted as \(_{n}C_{r}\) or \(\binom{n}{r}\). It is also a component of the binomial theorem, which expands expressions raised to a power.

To compute the value of a binomial coefficient like \(_{9}C_{5}\), we utilize the formula \(_{n}C_{r} = \frac{n!}{r!(n-r)!}\). Substituting the values of n and r, we can calculate the number of combinations. This concept is vital in probability and statistics, as it helps in determining the likelihood of certain outcomes.

Permutations and Combinations

Permutations and combinations are two fundamental concepts in combinatorics. They deal with the arrangement and selection of objects from a particular set. While both account for different scenarios, they are interconnected through the concepts of order.

Permutations

Permutations take into account the order of selection. In other words, the arrangement 'ABC' is different from 'CBA'. The number of permutations of n objects taken r at a time is denoted as \(_nP_r\) and is calculated using \(_nP_r = \frac{n!}{(n-r)!}\).

Combinations

On the other hand, combinations do not consider the order of the selection, meaning 'ABC' is the same as 'CBA'. The number of combinations is represented as \(_nC_r\), which we've seen in the earlier exercise \(_{9}C_{5}\). By understanding when to use permutations versus combinations, students can more accurately analyze problems involving probabilistic events and strategic decision making.