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Welcome to our Math lesson on Working with Binomial Coefficients and the Limitation of Pascal's Triangle, this is the fourth lesson of our suite of math lessons covering the topic of Binomial Expansion and Coefficients, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.
As we explained in the previous lesson, Pascal's Triangle is very helpful for determining the binomial coefficients when dealing with small degree binomials. For higher degree binomials, however, the use of Pascal's Triangle becomes more complicated, as the triangle's base widens. For example, we cannot use Pascal's Triangle to expand the binomial (x + y)317, as the base of Pascal's Triangle would contain 317 + 1 = 318 terms, which would need a lot of space to write out and is therefore not practical.
So it is necessary to use a more comprehensive method that is applicable in expressing the binomial coefficients of binomials of any degree. However, this method includes some concepts from Combinatorics - an area of mathematics primarily concerned with counting, both as a mean and as a tool for obtaining results, and certain properties of finite structures - the first thing to do is to explain the meaning of these concepts before continuing with the general formula of the binomial coefficients.
In mathematics, the factorial of a number is a particular function that multiplies that number by all natural numbers that are smaller than it. The symbol of factorial is the exclamation mark (!). For example,
and so on.
In general, we write for the factorial of any number n
Let's determine the value of 0! by using the following approach:
Therefore, not only 1! = 1 but also 0! = 1.
Following the procedure described earlier, we can write
This division is impossible, as a number cannot divide by zero because division by zero is undefined. Therefore, negative numbers cannot be written in the factorial form.
In mathematics, a combination is a concept used to describe the number of ways a group of elements extracted from a set of items can be combined with each other. More precisely, a combination is a way of selecting items from a collection (without repetition) where the order of selection does not matter. We denote a combination by C(n, k), where n is the total number of available items and k is the number of items per group.
For example, C(5, 3) indicates the number of possible three-items groups formed by a set of 5 elements. Thus, if we have a set of 5 elements, a, b, c, d and e, the number of three-letter combinations is 10 (abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde).
The general formula used to find the number of possible k-elements combinations in a set of n-elements is
Note: We will prove this formula (and many other related things) when dealing with Combinatorics as a separate chapter.
In the example above, we have n = 5 and k = 3, as the set contains 5 elements (a, b, c, d and e) and we are interested to form groups of three elements each. Therefore, applying the combinations formula
after the substitution of the known values we obtain
How many combinations are possible when forming groups of 4 balls out of a set of 9 balls?
From theory, it is known that the number of elements in the set is denoted by n. In our example, n = 9. On the other hand, the number of elements each group contains is denoted by k. Here, k = 4.
From the combinations formula, we have
Substituting the known values, we obtain for the number of possible 4-balls groups formed by a set of 9 balls is
We often express the (n, k) part not as a row but as a column instead. In this case, we no longer write the symbol C for combinations. In other word, the scripts below are equivalent.
This form is used to express the Binomial Coefficients Theorem that we are going to present below.
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