Problem 14 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the given expression. $$ C(9,3) $$

Step-by-Step Solution

Answer

C(9,3) can be calculated using the binomial coefficient formula: \[C(n, k) = \frac{n!}{k!(n-k)!}\]. Plugging in the values, we get: \[C(9,3) = \frac{9!}{3!(9-3)!} = \frac{362,880}{6 \times 720} = 84\]. Thus, C(9,3) equals 84.

Step 1: Identify the values of n and k

First, we need to identify the values of n and k in the given expression. We are given C(9,3), so n = 9 and k = 3.

Step 2: Compute the factorial of n, k, and (n-k)

Next, we will compute the factorials of n, k, and (n-k): \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880\) \(3! = 3 \times 2 \times 1 = 6\) \((9 - 3) = 6\) and \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)

Step 3: Use the binomial coefficient formula

Now, we will use the binomial coefficient formula to find the value of C(9,3): \[C(9,3) = \frac{9!}{3!(9-3)!} = \frac{362,880}{6 \times 720} = \frac{362,880}{4,320}\]

Step 4: Simplify the expression

Finally, we will simplify the expression: C(9,3) = 84 Therefore, the binomial coefficient C(9,3) is equal to 84.

Key Concepts

Factorial

The concept of a factorial is fundamental in combinatorics and is represented by an exclamation mark (!). For a non-negative integer 'n', the factorial of 'n' (denoted as 'n!') is the product of all positive integers less than or equal to 'n'.

In formal terms,
\[ n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \]
For example, '5!' equals to
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
There is a special case where the factorial of 0 is defined to be 1. This might seem odd, but it's a convenient convention when calculating combinations and permutations.

Factorials grow very quickly with increasing values of 'n'. Consequently, the computation of large factorials requires special algorithms or software functions that efficiently handle large numbers. They are not only used in binomial coefficients but also in permutations, power series, and other areas of mathematics.

Combinations

Combinations refer to the way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. When we calculate combinations, we are essentially looking at the different ways we can choose a subset of items from a larger set.

The formula for combinations is given by the binomial coefficient, which is a function of two numbers, 'n' and 'k', and is denoted as 'C(n, k)' or '\( \binom{n}{k} \)'. The formula for 'C(n, k)' is:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
where 'n!' is the factorial of 'n', 'k!' is the factorial of 'k', and '(n-k)!' is the factorial of 'n-k'.

It's important to note that 'k' must be an integer equal to or less than 'n'. In our example with 'C(9,3)', the question was how many ways can we choose 3 items from a set of 9, and the answer is 84 different ways.

Permutations

Permutations are arrangements of items where order does matter. While combinations focus on the selection of items, permutations are concerned with the arrangement of these selected items.

The number of permutations of 'n' objects taken 'k' at a time is given by the formula:
\[ P(n, k) = \frac{n!}{(n-k)!} \]
If you are arranging all 'n' items, the formula simplifies to just 'n!'. For example, the number of permutations of 5 different books is \(5! = 120\) different ways to arrange these books.

Permutations are essential in fields where order is significant, like coding, scheduling problems, and the creation of passwords. Recognizing when to use permutations versus combinations is crucial in solving problems related to probabilities and arrangements.