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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