## Permutations and Combinations

### 3.5 The Binomial Theorem

Lesson materials located below the video overview.

When I was in school we were required to memorize the following expansions:

\(\left( x + y \right)^{2} = x^{2} + 2xy + y^{2}\)

\( \left( x + y \right)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3}\)

\( \left( x + y \right)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}\)

We could add to this:

\(\left( x + y \right)^{0} = 1\)

\(\left( x + y \right)^{1} = x +y\)

\(\left( x + y \right)^{2} = x^{2} + 2xy + y^{2}\)

\( \left( x + y \right)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3}\)

\( \left( x + y \right)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}\)

Do you see Pascal’s triangle in the coefficients? Isn’t that bizarre?

Let’ s explain why this is so.

In expanding brackets one selects one term from each set of parentheses and makes sure to collect all possible combinations. For example:

\(\left(a+b)\right)\left(x+y\right) = ax + bx + ay + by\)

\(\left(w+p+q\right)\left(3+q\right) = 3w + 3p + 3q + wq + pq + q^{2}\)

\(\left(a + b + c\right)\left(x + y\right)\left(p+q+r\right)\left(s+t+u+v\right) = axps + ayqu + cxps + \cdots\)

Imagine expanding the next quantity on the list:

\(\left(x + y\right)^{5} = (x+y)(x+y)(x+y)(x+y)(x+y)\)

Is there any reason to believe we will see the coefficients of the fifth row of Pascal’s triangle?

The term \(x^{5}\) will appear once by choosing “\(x\)” from each set of parentheses.

The term \(x^{4}y\) will appear five times:

once by choosing *x, x, x ,x *and then *y*.

once by choosing *x, x, x, y*, and then *x*.

once by choosing *x, x, y, x*, and then *x*.

once by choosing *x, y, x, x*, and then *x*.

once by choosing *y, x, x, x*, and then *x*.

That is, \(x^{4}y\) will appear the same number of times as it is possible to arrange four \(x\)s and one \(y\). This can be done \(\dfrac{5!}{4!1!}=5\) ways, which is the entry of Pascal’s triangle on the fifth row, four places in from the left and one place in from the right.

The term \(x^{3}y^{2}\) will appear as many times as it is possible to arrange three \(x\)s and two \(y\)*s*, that is, \(\dfrac{5!}{3!2!} = 10\) times.

The term \(x^{2}y^{3}\) appears ten times, the term \(xy^{4}\) five times, and the term \(y^{5}\) once. We have:

\( \left( x + y \right)^{5} = x^{5} + 5x^{4}y + 10x^{3}y^{2} + 10x^{2}y^{3} + 5xy^{4} + y^{5}\)

with the numbers 1, 5, 10, 10, 5, 1 indeed the entries of the fifth row of Pascal’s triangle.

In fact, in general we have:

\( \left( x + y \right)^{n} = x^{n} + \cdots + \dfrac{(a+b)!}{a!b!}x^{a}y^{b} + \cdots + y^{n}\)

with the coefficient of \(x^{a}y^{b}\) just the number of ways to arrange \(a\) \(x\)s and \(b\) \(y\) s.

Let’s have some fun with the formulas:

\(\left( x + y \right)^{0} = 1\)

\(\left( x + y \right)^{1} = x +y\)

\(\left( x + y \right)^{2} = x^{2} + 2xy + y^{2}\)

\( \left( x + y \right)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3}\)

\( \left( x + y \right)^{4} = x^{4} + 4x^{3}y + 4x^{2}y^{2} + 4xy^{3} + y^{4}\)

\( \left( x + y \right)^{5} = x^{5} + 5x^{4}y + 10x^{3}y^{2} + 10x^{2}y^{3} + 5xy^{4} + y^{5}\)

and so on.

** **** **

**1. **Put \(x=10\) and \(y=1\). Notice that, for instance:

\(11^{4} = \left(10 + 1\right)^{4} = 10^{4} + 4 \cdot 10^{3} + 6 \cdot 10^{2} + 4 \cdot 10 + 1 = 10000 + 4000 + 600 + 40 + 1 = 14641\).

This explains the connection of the powers of 11.

** **

** **

**2. **Put \(x=1\) and \(y = 1\). Notice that, for instance:

\(2^{4} = \left(1 + 1\right)^{4} = 1^{4} + 4 \cdot 1^{3} + 6 \cdot 1^{2} + 4 \cdot 1 + 1 = 1 + 4 + 6 + 4 + 1\).

This explains, again, why the sum of entries in a row of entries of Pascal’s triangle is a power of two.

**3. **Put \(x =1\) and \( y = -1\). Notice that, for instance:

\(0 = \left(1 – 1\right)^{4} = 1^{4} + 4 \cdot 1^{3} \cdot (-1) + 6 \cdot 1^{2} \cdot (-1)^{2} + 4 \cdot 1 \cdot (-1)^{3} + 1 \cdot (-1)^{4} = 1 – 4 + 6 – 4 + 1\).

This explains, again, why the alternating sum of entries in a row of Pascal’s triangle is always zero.

Practice: Take any row of Pascal’s triangle and multiply its entries by consecutive powers of 2 taken in reverse order. Prove that their sum is sure to be a power of 3. (For example:\(1 \times 16 + 4 \times 8 + 6 \times 4 + 4 \times 2 + 1 \times 1 = 81 = 3^{4}\) .) |

Stated formally, we have …

Binomial Theorem:\(\left(x + y\right)^{n} = x^{n} + \dfrac{n!}{(n-1)!1!}x^{n-1}y+\dfrac{n!}{(n-2)!2!}x^{n-2}y^{2} + \cdots + \dfrac{n!}{a!b!}x^{a}y^{b} + \cdots + y^{n}\)The coefficients are the entries of the \(n\)th row of Pascal’s triangle. |

**A COMMENT ON NOTATION: **

Mathematicians use the notation \( n \choose a b\) for the expression \(\dfrac{n!}{a!b!}\) with \(a+b=n\) . Thus the binomial theorem can be written:

\(\left(x + y\right)^{n} = {n \choose n 0} x^{n} + {n \choose n-1 1}x^{n-1}y+{n \choose n-2 2}x^{n-2}y^{2} + \cdots + {n \choose a b} x^{a}y^{b} + \cdots + {n \choose 0 n}y^{n}\)

Actually, mathematicians usually suppress one of the terms in the notation and write just \({n \choose a}\) for \({ n \choose a b}\). (It is clear what the missing term must be.) For example, \({7 \choose 5} = {7 \choose 5 2} = \dfrac{7!}{5!2!} \). Thus the binomial theorem might be written more compactly as:

\(\left(x + y\right)^{n} = {n \choose n} x^{n} + {n \choose n-1}x^{n-1}y+{n \choose n-2}x^{n-2}y^{2} + \cdots + {n \choose a} x^{a}y^{b} + \cdots + {n \choose 0}y^{n}\)

The entries of Pascal’s triangles, \({n \choose a}\) , are also called *binomial coefficients* because of this connection to the binomial theorem.

See exercise 40 in the next section for the “trinomial theorem,” and beyond!

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