Problem 51 - College Algebra

Question

Use the formula for \(_{n} C_{r}\) to solve Of 12 possible books, you plan to take 4 with you on vacation. How many different collections of 4 books can you take?

Step-by-Step Solution

Answer

The number of different collections of 4 books that you can take from a set of 12 is 495.

Step 1: Define Values

In this case, we are selecting 4 books from a set of 12. So, our total number of items \(n\) is 12 and the number of items to choose \(r\) is 4.

Step 2: Use the Combination Formula

We apply the formula for combinations which is: \(_{n} C_{r} = \frac{n!}{r!(n-r)!}\). Substituting \(n = 12\) and \(r = 4\) we have \(_{12}C_{4} = \frac{12!}{4!(12-4)!}\).

Step 3: Evaluate the Factorials

Before simplifying, evaluate the factorials. Factorial means multiplying all positive integers up to that number. So, \(12! = 12*11*10*9*8*7*6*5*4*3*2*1, 4! = 4*3*2*1, 8! = 8*7*6*5*4*3*2*1\). Substitute these back into the equation.

Step 4: Simplify the Equation

Now, simplify the equation. Many of the terms will cancel out, leaving only: \(_{12}C_{4} = \frac{12*11*10*9}{4*3*2*1}\)

Step 5: Calculate the final result

Finally, solve the equation to find our answer: \(_{12}C_{4} = 495\)

Key Concepts

Factorials

In mathematics, the concept of factorials is essential when working with combinations and permutations. A factorial, denoted by the symbol !, represents the product of all positive integers from 1 up to a given number. For example, "5!" reads "5 factorial" and is calculated as 5 \times 4 \times 3 \times 2 \times 1. Factorials are used primarily to simplify expressions involving permutations and combinations.
\[n! = n \times (n-1) \times (n-2) \times \cdots \times 1\]
They grow very quickly with increasing values of n. For instance, 10! represents a much larger number than 5!

• 1! = 1
• 2! = 2 \times 1 = 2
• 3! = 3 \times 2 \times 1 = 6
• 4! = 24
• 5! = 120

Factorials help determine the total number of possible ways to arrange a given set of items, which is paramount in solving problems involving combinations and permutations.

Combination Formula

The combination formula plays a critical role when you need to find the number of ways to select items from a set without regard to the order. This formula is different from permutations where order does matter. The general combination formula is given by:\[ _{n}C_{r} = \frac{n!}{r!(n-r)!}\]In this formula, "n" represents the total number of items to choose from, while "r" is the number of items to choose. **The key here is that the order of selection does not matter**. This is why we only take unique selections into account.
For instance, in the textbook problem, "n" is 12, as there are 12 books, and "r" is 4, since we choose 4 books. Substituting these values into the formula gives us \(_{12}C_{4} = \frac{12!}{4! \times (12-4)!}\). This allows us to find out how many unique sets of 4 books you can take from the 12 available ones.
Using the combination formula effectively provides solutions to many problems involving groups and selections in statistics and probability.

Binomial Coefficients

Binomial coefficients are a particular set of numbers that occur frequently in mathematics and play an integral part in combinatorics and algebra. They are represented by the notation: \( _{n}C_{r}\) and are synonymous with the combination formula discussed previously. These coefficients arise in the binomial theorem, which expresses the expansion of a binomial raised to a power.
The binomial theorem is important for expanding powers of sums and is articulated as follows:

• \((x+y)^{n} = \sum_{r=0}^{n} {_n}C_{r} x^{(n-r)} y^{r}\)

Here, each term in the expansion involves a binomial coefficient \({_{n}}C_{r}\), reflecting the number of ways to choose r elements from a total of n. Binomial coefficients ensure that each combination of terms is appropriately weighted. For instance, the textbook example utilizes this concept; finding \(_{12}C_{4}\) is essentially computing a binomial coefficient.
Understanding binomial coefficients is fundamental for comprehending probability distributions, algebraic identities, and in solving complex combinatorial problems.