Pascal's Triangle


What is Pascal's Triangle?

Pascal's triangle is a triangle-shaped grid of numbers, defined by a special pattern of related sums.The general shape of the triangle isand it is built from rows of numbers, each having one more member than its predecessor. In other words, the $n$-th row from the top has $n$ terms.The top of the triangle starts with two rows consisting of one $1$ in the first row and two $1$'s: in the second row.From here on out, the next row will always be obtained via the following two rules:1) The first and last number in the row will be $1$.2) Every other number will be the sum of the two numbers above it.This tells us that each number that is not first or last must be spaced out in between the numbers above it. It also suggests that we could just keep going forever. We could of course, but when you need to use Pascal's triangle for a practical purpose, it's incredibly unlikely that you'll need more than ten rows (most of the time six or fewer, in my experience).Let's create row three.Start with the $1$'s on the end.The third row has three terms total and we've already got two. The middle term will be the sum of the two numbers above it on either side.Row four will have four terms. Once again we will start and end with $1$, and find the other two by adding the above-adjacent numbers:See if you can recreate rows five and six before scrolling down!

Uses of Pascal's Triangle

There are two common uses of Pascal's triangle, which each stem from the same operation: combinations » and binomial expansions ».Combinations is a counting mechanism in combinatorics that tells you how many different groups you can select when choosing a subset of a larger group. For example, if you are told to mix exactly two colors of paint together in art class, and you have eight colors available to choose from, then the total number of ways you can make your selection is $8 \choose 2$ (which is read “ten choose two”).Every row of Pascal's Triangle gives the answers to a specific set of combinations (except the first row of the triangle that consists of a lone $1$). If you have $n$ total objects to select from, then look in row $n+1$.Examine Pascal's Triangle again and look at row nine.Row nine shows the counts for number of ways to choose subsets of eight objects.Why isn't row eight the right row for eight object combinations? Because choosing various subsets of eight objects can be done nine ways: you could choose $0$ objects, $1$ object, $2$ objects, all the way up to $8$ objects.Scroll along the row and see what those numbers mean to us:There is $1$ way to choose $0$ objects out of $8$ (makes sense).There are $8$ ways to choose $1$ object out of $8$ (also makes sense intuitively).There are $28$ ways to choose $2$ objects out of $8$ (harder to digest with intuition, but true).There are $56$ ways to choose $3$ objects out of $8$ (harder still).There are $70$ ways to choose $4$ objects out of $8$ (that's a lot!).There are $56$ ways to choose $5$ objects out of $8$ (same as three, think about why!).And the pattern continues: we would see that there are $28$ ways to choose $6$ objects out of $8$, $8$ ways to choose $7$ objects out of $8$ available (makes sense because there are $8$ ways you could choose $1$ object to exclude) and only $1$ way to select all $8$ objects.In this way, Pascal's Triangle allows us to calculate what is otherwise a more tedious calculation (see the full lesson on combinations and permutations » for a more in-depth look at counting ways to make choices).Here's another example - one small enough that we can look at all possibilities and verify that our answers match Pascal's Triangle.Combinations ExampleYour family has decided to donate to up to four charities this year, charities A, B, C, and D. What are all the possible ways your family could choose to spread its donation among the four?To answer this question, we would first need to choose how many charities to donate to, and then which ones.1) Your family could choose to donate to all four.2) Your family could choose to donate to three of them.3) Your family could choose to donate to two of them.4) Your family could choose to donate to only one of them.5) Your family could change its mind and make zero donations this year (that doesn't seem very nice but we must include it as a possibility, since it could happen).Now let's count the ways each scenario could happen, manually.Donate to all four - $1$ way to do this:$$\{A,B,C,D\}$$Donate to three charities - $4$ ways to do this:$$ \begin{align} \{A,B,C\} \\ \{A,B,D\} \\ \{A,C,D\} \\ \{B,C,D\} \end{align}$$Donate to two charities: $6$ ways to do this:$$ \begin{align} \{A,B\} \\ \{A,C\} \\ \{A,D\} \\ \{B,C\} \\ \{B,D\} \\ \{C,D\} \end{align}$$Donate to one charity: $4$ ways to do this:$$ \begin{align} \{A\} \\ \{B\} \\ \{C\} \\ \{D\} \end{align}$$Donate to no charities: $1$ way to do this:$$\{\}$$(The Null Set)So we have counted by hand to determine that ${4 \choose 4} = 1$, ${4 \choose 3} = 4$, ${4 \choose 2} = 6$, ${4 \choose 1} = 4$, and ${4 \choose 0} = 1$.These results match row five of Pascal's triangle.

Binomial Expansions

(Repeated here from its own Misfit Math post »)By algebraic definition, binomials are the sum of two objects. Let $(x+y)$ be any binomial (that is to say, whatever the two objects being added together are, let $x$ be the first and $y$ be the second).Binomial expansion refers to the result you get when you raise a binomial to a power. As much as we may wish it so,$$(x+y)^n \ne x^n + y^n$$This can be instantly proven for $n=2$:$$(x+y)^2 = x^2 + 2xy + y^2 \ne x^2 + y^2$$The hard truth is that, in order to multiply binomials, you need to distribute (FOIL, in the case of $n=2$). That is, for example,$$(x+y)^2 = (x+y)(x+y)$$This quickly becomes tedious and annoying - even multiplying three binomials together seems like a drag.$$(x+y)^3 = (x+y)(x+y)(x+y)$$However, Pascal's Triangle gives a magnificent shortcut for this process! If you want to multiply $(x+y)^n$, follow these guidelines:
  • The variable part of the first term is $x^n$
  • The variable part of the next term is $x^{n-1} y$
  • Each sequential term's variable piece decreases the exponent on $x$ and increases the exponent on $y$. i.e. the third term is of the form $x^{n-2} y^2$
  • Keep going until you have decreased $x$ to nothing, at which point your last term will be $y^n$
  • If you have $(x+y)$ then all terms are positive, and if you have $(x-y)$ then the first term is positive, the second term is negative, and each term after that alternates
  • Finally, and significantly - the coefficients of each term will come from the $n+1$ row in Pascal's Triangle
Here are three examples. See if you can line up the guidelines above with the results. When you know the pattern, you don't actually have to do the tedious multiplication to get the answer!$$(x+y)^5$$
$$(2a + 3b)^4$$