Problem 83 - College Algebra

Question

Prove that $$\left(\begin{array}{l}n \\\r\end{array}\right)-\left(\begin{array}{c}n \\\n-r\end{array}\right)$$.

Step-by-Step Solution

Answer

The binomial coefficients \( \left(\begin{array}{l}n \ \end{array}\right) \) and \( \left(\begin{array}{c}n \ -r\end{array}\right) \) are equal because their calculations result in the same value: \( \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!} \).

Step 1 - Calculate the left-hand binomial coefficient

Firstly, let's calculate the left-hand side binomial coefficient \( \left(\begin{array}{l}n \ \end{array}\right) \) using the formula \( \left(\begin{array}{l}n \ \end{array}\right) = \frac{n!}{r!(n-r)!} \). So, it equals: \(\frac{n!}{r!(n-r)!}\).

Step 2 - Calculate the right-hand binomial coefficient

Secondly, calculate the right-hand binomial coefficient \( \left(\begin{array}{c}n \ -r\end{array}\right) \) using the formula \( \left(\begin{array}{c}n \ -r\end{array}\right) = \frac{n!}{(n-r)!(n-(n-r))!} \). This simplifies to: \( \frac{n!}{(n-r)!r!} \).

Step 3 - Compare the two binomial coefficients

Now that both coefficients have been calculated, it's clear that \( \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!} \) which shows that \( \left(\begin{array}{l}n \ \end{array}\right) \) is indeed equal to \( \left(\begin{array}{c}n \ -r\end{array}\right) \).

Key Concepts

Combination

When solving mathematical problems, the term 'combination' refers to a selection of items from a larger set, such that the order of selection does not matter. In simpler terms, it's like picking a few fruits from a basket without caring which fruit you grab first. For instance, if you have a set of letters \(A, B, C\), the combinations of taking 2 at a time (\

Factorial Notation

If you ever wondered how mathematicians calculate the total number of ways to arrange a specific number of items, they often use something called factorial notation. It's represented by an exclamation mark \( ! \) and is a way to express the product of an integer and all the integers below it down to 1. In essence, it counts the number of ways to arrange items in a sequence. For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\). A crucial aspect of factorial notation is that by definition, \(0! = 1\), which might seem strange at first but is very useful in mathematical proofs and formulas. When dealing with combinations or permutations, factorial notation makes it feasible to calculate large numbers without having to multiply each digit individually. For instance, determining the number of ways to arrange 10 books on a shelf would be a cumbersome task without the shorthand provided by factorial notation.

Permutation

In contrast to combinations, permutation is all about order. It's a way of arranging objects where the sequence is important, like when you're trying to figure out possible line-ups for a team or the different ways you can arrange your favorite songs in a playlist. Imagine lining up your three favorite desserts; whether the pie comes before or after the cake makes a difference in a permutation.

For numerically calculating permutations, we use a similar notation to combinations but with a subtle difference that emphasizes order. The formula for permutations when you want to arrange \( r \) items out of \( n \) available is given as \( P(n, r) = \frac{n!}{(n-r)!} \). This formula tells us the number of different ways to arrange \( r \) items from a set of \( n \), and it clearly shows us the difference from combinations where the order doesn't matter. Understanding how these principles interplay is fundamental in solving more complex probability, statistics, and combinatorics problems.