Factorials
A factorial is a function that multiplies a whole number by every positive whole number below it.
For any non-negative integer n, the factorial of n, written as n!, is defined as:
n! = n × (n - 1) × (n - 2) × … × 3 × 2 × 1
Example
Fred knows that the code to his locker is made up of the digits 5, 6, 7 and 8, but he can’t remember the order. How many different possible codes exist?
There are 4 choices for the first digit:
5*** 6*** 7*** 8***
There are then 3 choices for the second digit:
| 56** | 65** | 75** | 85** |
| 57** | 67** | 76** | 86** |
| 58** | 68** | 78** | 87** |
This leaves 2 choices for the third digit:
| 567* | 657* | 756* | 856* |
| 568* | 658* | 758* | 857* |
| 576* | 675* | 765* | 865* |
| 578* | 678* | 768* | 867* |
| 586* | 685* | 785* | 875* |
| 587* | 687* | 786* | 876* |
Finally, only one digit remains for the last position:
5678 6578 7568 8567
5687 6587 7586 8576
5768 6758 7658 8657
5786 6785 7685 8675
5867 6857 7856 8756
5876 6875 7865 8765
There are 24 possible codes.
\( 4 \times 3 \times 2 \times 1 = 24 \)
Some useful values:
\[
0! = 1,\quad
1! = 1,\quad
2! = 2,\quad
3! = 6,\quad
4! = 24,\quad
5! = 120
\]
In general:
\( n! = n(n-1)(n-2)\cdots 3\cdot 2\cdot 1 \)
Useful identities:
\( n! = n\,(n-1)! \)
\( (n-r+1)(n-r)! = (n-r+1)! \)
\[
(n+1)!
= (n+1)\bigl((n+1)-1\bigr)\bigl((n+1)-2\bigr)\bigl((n+1)-3\bigr)\,\ldots\,3\times 2\times 1
\]
\[
= (n+1)\,n\,(n-1)\,(n-2)\,\ldots\,3\times 2\times 1
\]
\[
= (n+1)\,n!
\]
\[
= (n+1)\,n\,(n-1)!
\]
\[
(n-2)!
= (n-2)\bigl((n-2)-1\bigr)\bigl((n-2)-2\bigr)\bigl((n-2)-3\bigr)\,\ldots\,3\times 2\times 1
\]
\[
= (n-2)\,(n-3)\,(n-4)\,(n-5)\,(n-6)\,\ldots\,3\times 2\times 1
\]
\[
= (n-2)\,(n-3)!
\]
Permutations and Combinations
\[
\text{Arrange } r \text{ objects from a total of } n
\]
\( {}^{n}P_{r} = \frac{n!}{(n-r)!} \)
\[
\text{Select } r \text{ objects from a total of } n
\]
\( {}^{n}C_{r} = \frac{n!}{r!(n-r)!} \)
Example
There are 56 ways of choosing 5 tins from 8 brands, but 6720 ways of ordering them.
\[
{}^{8}C_{5}
= \frac{8!}{5!\,(8-5)!}
\]
\[
= \frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{5!\,3!}
\]
\[
= \frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}
{(5\times 4\times 3\times 2\times 1)(3\times 2\times 1)}
\]
\[
= \frac{8\times 7\times 6}{3\times 2\times 1}
\]
\[
= \frac{336}{6}
\]
\[
= 56
\]
\[
{}^{8}P_{5}
= \frac{8!}{(8-5)!}
\]
\[
= \frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3!}
\]
\[
= \frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1}
\]
\[
= 8\times 7\times 6\times 5\times 4
\]
\[
= 6720
\]
Properties of combinations
\[
{}^{n}C_{n}
= \frac{n!}{n!\,(n-n)!}
\]
\[
= \frac{n!}{n!\,0!}
\]
\[
= \frac{n!}{n!}
\]
\[
= 1
\]
\[
{}^{n}C_{0}
= \frac{n!}{0!\,(n-0)!}
\]
\[
= \frac{n!}{0!\,n!}
\]
\[
= \frac{n!}{n!}
\]
\[
= 1
\]
\[
{}^{n}C_{\,n-r}
= \frac{n!}{(n-r)!\,\bigl(n-(n-r)\bigr)!}
\]
\[
= \frac{n!}{(n-r)!\,(n-n+r)!}
\]
\[
= \frac{n!}{(n-r)!\,r!}
\]
\[
= \frac{n!}{r!\,(n-r)!}
\]
\[
= {}^{n}C_{\,r}
\]
Pascal’s Triangle
The triangle is built by adding the two numbers above.
This can be written as \( {}^{n}C_{r} \), where n is the row and r is the column.
Notice how \({}^{3}C_{3} = {}^{3}C_{0} = 1\).
\({}^{3}C_{2} = {}^{3}C_{1} = 3\).
\({}^{4}C_{3} = {}^{4}C_{1} = 4\).
\[
{}^{n}C_{r} + {}^{n}C_{\,r+1}
= {}^{\,n+1}C_{\,r+1}
\]
\[
\text{e.g.}\quad
{}^{4}C_{2} + {}^{4}C_{3}
= 10
= {}^{5}C_{3}
\]
\[
{}^{n}C_{r-1} + {}^{n}C_{r}
= {}^{\,n+1}C_{r}
\]
\[
\text{e.g.}\quad {}^{4}C_{1} + {}^{4}C_{2}
= 10
= {}^{5}C_{2}
\]
Binomial Coefficients
The coefficients in the expansion of \( (a+b)^n \) are called binomial coefficients.
\( {}^{n}C_{r} = \frac{n!}{r!(n-r)!} \)
Can be written
\[
\begin{pmatrix}
n \\[4pt]
r
\end{pmatrix}
\\
\begin{pmatrix}
\text{Row} \\[4pt]
\text{Column}
\end{pmatrix}
\]
where n is the row and r is the column in pascal's triangle
Example
\[
{}^{4}C_{2}\ \text{can be written}\
\begin{pmatrix}
4 \\[4pt]
2
\end{pmatrix}
\]
\[
{}^{4}C_{2}
= \frac{4!}{2!\,(4-2)!}
\]
\[
= \frac{4!}{2!\,2!}
\]
\[
= \frac{4\times 3\times 2!}{2!\,2!}
\]
\[
= \frac{4\times 3}{2!}
\]
\[
= \frac{12}{2} = 6
\]
Properties to learn
\[
\begin{pmatrix}
n \\[4pt]
r
\end{pmatrix}
+
\begin{pmatrix}
n \\[4pt]
r+1
\end{pmatrix}
=
\begin{pmatrix}
n+1 \\[4pt]
r+1
\end{pmatrix}
\]
\[
\begin{pmatrix}
n \\[4pt]
r-1
\end{pmatrix}
+
\begin{pmatrix}
n \\[4pt]
r
\end{pmatrix}
=
\begin{pmatrix}
n+1 \\[4pt]
r
\end{pmatrix}
\]
\[
\begin{pmatrix}
n \\[4pt]
n-r
\end{pmatrix}
=
\begin{pmatrix}
n \\[4pt]
r
\end{pmatrix}
\]
\[
\begin{pmatrix}
n \\[4pt]
n
\end{pmatrix}
=
\begin{pmatrix}
n \\[4pt]
0
\end{pmatrix}
= 1
\]
Example
Show that
\(
\begin{pmatrix}
n+1 \\[4pt]
4
\end{pmatrix}
-
\begin{pmatrix}
n \\[4pt]
4
\end{pmatrix}
=
\begin{pmatrix}
n \\[4pt]
3
\end{pmatrix}
\) , where \( n \ge 4 \)
Start by writing out the left-hand side in the form \( {}^{n}C_{r} = \frac{n!}{r!(n-r)!} \)
\[
\begin{pmatrix}
n+1 \\[4pt]
4
\end{pmatrix}
-
\begin{pmatrix}
n \\[4pt]
4
\end{pmatrix}
=
\frac{(n+1)!}{4!\,\bigl((n+1)-4\bigr)!}
-
\frac{n!}{4!\,(n-4)!}
\]
\[
=
\frac{(n+1)!}{4!\,(n-3)!}
-
\frac{n!}{4!\,(n-4)!}
\]
multiplying the second term by 1 in the form of (n-3)/(n-3) gives
\[
=
\frac{(n+1)!}{4!\,(n-3)!}
\;-\;
\frac{n!}{4!\,(n-4)!}\times 1
\]
\[
=
\frac{(n+1)!}{4!\,(n-3)!}
\;-\;
\frac{n!}{4!\,(n-4)!}
\times
\frac{(n-3)}{(n-3)}
\]
\[
=
\frac{(n+1)!}{4!\,(n-3)!}
\;-\;
\frac{n!\,(n-3)}{4!\,(n-3)!\,(n-4)!}
\]
which reduces to
\[
=
\frac{(n+1)!}{4!\,(n-3)!}
\;-\;
\frac{n!\,(n-3)}{4!\,(n-3)!}
\]
since (n-3)!= (n-3)(n-4)!
Now the denominators are the same
\[
=
\frac{(n+1)! \;-\; n!\,(n-3)}{4!\,(n-3)!}
\]
take out common factor n!
\[
=
\frac{
n!\,\bigl[(n+1) - (n-3)\bigr]
}{
4!\,(n-3)!
}
\]
\[
=
\frac{
n!\,\bigl[1 + 3\bigr]
}{
4!\,(n-3)!
}
\]
\[
=
\frac{
n!\,4
}{
4!\,(n-3)!
}
\]
\[
=
\frac{n!\,4}{4 \times 3!\,(n-3)!}
\]
\[
=
\frac{n!}{3!\,(n-3)!}
\]
\[
=
\begin{pmatrix}
n \\[4pt]
3
\end{pmatrix}
\]
Equations
Example
Solve the equation
\(
\begin{pmatrix}
n \\[4pt]
n-2
\end{pmatrix}
= 55
\)
From the properties above, since
\[
\begin{pmatrix}
n \\[4pt]
n-r
\end{pmatrix}
=
\begin{pmatrix}
n \\[4pt]
r
\end{pmatrix}
\]
\[
\begin{pmatrix}
n \\[4pt]
n-2
\end{pmatrix}
=
\begin{pmatrix}
n \\[4pt]
2
\end{pmatrix}
\]
Solution
\[
\begin{pmatrix}
n \\[4pt]
n-2
\end{pmatrix}
= 55
\]
\[
\therefore\;
\begin{pmatrix}
n \\[4pt]
2
\end{pmatrix}
= 55
\]
\[
\text{Look for 55 in column 2 of Pascal’s triangle.}
\]
\[
\text{This occurs in row 11.}
\]
\[
n = 11
\]
\[
\frac{n!}{2!\,(n-2)!} = 55
\]
\[
\frac{n(n-1)(n-2)(n-3)\,\dots\,3 \times 2 \times 1}{2!\,(n-2)!} = 55
\]
\[
\frac{n(n-1)(n-2)!}{2!\,(n-2)!} = 55
\]
\[
\frac{n(n-1)}{2!} = 55
\]
tidy up
\[
\frac{n(n-1)}{2 \times 1} = 55
\]
\[
\frac{n(n-1)}{2} = 55
\]
make equation and solve
\[
n(n-1) = 110
\]
\[
n^{2} - n = 110
\]
\[
n^{2} - n - 110 = 0
\]
\[
(n - 11)(n + 10) = 0
\]
\[
\text{so}
\]
\[
n = 11 \quad \text{or} \quad n = -10
\]
\[
\text{discard } n = -10 \text{ since } n > 0
\]
\[
n = 11
\]
Example
Solve the equation
\( \begin{pmatrix} n \\[4pt] n-3 \end{pmatrix} = 84 \)
\[
\begin{pmatrix}
n \\[4pt]
n-3
\end{pmatrix}
= 84
\]
\[
\therefore\;
\begin{pmatrix}
n \\[4pt]
3
\end{pmatrix}
= 84
\]
\[
\text{Look for 84 in column 3 of Pascal’s triangle.}
\]
\[
\text{This occurs in row 9.}
\]
\[
n = 9
\]
Binomial Expansions
\[
(a+b)^{n}
=
C^{n}_{0}\,a^{n}b^{0}
+
C^{n}_{1}\,a^{\,n-1}b^{1}
+
C^{n}_{2}\,a^{\,n-2}b^{2}
+\;\dots\;+\;
C^{n}_{r}\,a^{\,n-r}b^{r}
+\;\dots\;+\;
C^{n}_{n}\,a^{\,0}b^{n}
\]
Which can be written
\[
(a+b)^{n}
=
\binom{n}{0}a^{\,n}b^{\,0}
+
\binom{n}{1}a^{\,n-1}b^{\,1}
+
\binom{n}{2}a^{\,n-2}b^{\,2}
+\;\dots\;+\;
\binom{n}{r}a^{\,n-r}b^{\,r}
+\;\dots\;+\;
\binom{n}{n}a^{\,0}b^{\,n}
\]
In general:
\( (a+b)^n = \sum_{r=0}^{n} {}^{n}C_{r} a^{\,n-r} b^{\,r} \)
\[
=
\sum_{r=0}^{n}
\begin{pmatrix}
n \\[4pt]
r
\end{pmatrix}
a^{\,n-r} b^{\,r}
\]
Example
\[
(x+y)^{5}
=
1\,x^{5}y^{0}
+5\,x^{4}y^{1}
+10\,x^{3}y^{2}
+10\,x^{2}y^{3}
+5\,x^{1}y^{4}
+1\,x^{0}y^{5}
\]
\[
=
C^{5}_{0}x^{5}y^{0}
+
C^{5}_{1}x^{4}y^{1}
+
C^{5}_{2}x^{3}y^{2}
+
C^{5}_{3}x^{2}y^{3}
+
C^{5}_{4}x^{1}y^{4}
+
C^{5}_{5}x^{0}y^{5}
\]
\[
=
\begin{pmatrix}
5 \\[4pt] 0
\end{pmatrix} x^{5}y^{0}
+
\begin{pmatrix}
5 \\[4pt] 1
\end{pmatrix} x^{4}y^{1}
+
\begin{pmatrix}
5 \\[4pt] 2
\end{pmatrix} x^{3}y^{2}
+
\begin{pmatrix}
5 \\[4pt] 3
\end{pmatrix} x^{2}y^{3}
+
\begin{pmatrix}
5 \\[4pt] 4
\end{pmatrix} x^{1}y^{4}
+
\begin{pmatrix}
5 \\[4pt] 5
\end{pmatrix} x^{0}y^{5}
\]
\[
=
x^{5}
+5x^{4}y
+10x^{3}y^{2}
+10x^{2}y^{3}
+5xy^{4}
+y^{5}
\]
Example
\[
(1+2x)^{3}
=
\begin{pmatrix}3\\0\end{pmatrix}1^{3}(2x)^{0}
+
\begin{pmatrix}3\\1\end{pmatrix}1^{2}(2x)^{1}
+
\begin{pmatrix}3\\2\end{pmatrix}1^{1}(2x)^{2}
+
\begin{pmatrix}3\\3\end{pmatrix}1^{0}(2x)^{3}
\]
\[
= 1 + 3(2x) + 3(2x)^{2} + (2x)^{3}
\]
\[
= 1 + 6x + 3\cdot 4x^{2} + 8x^{3}
\]
\[
= 1 + 6x + 12x^{2} + 8x^{3}
\]
\[
\text{The sum of the numbers in Pascal’s triangle is equal to } 2^{n}, \text{ where } n \text{ is the row number.}
\]
\[
\qquad\qquad
(a+b)^{n}
=
\sum_{r=0}^{n}
C^{n}_{r}\,a^{\,n-r} b^{\,r}
\]
\[
\text{when } a=1 \text{ and } b=1
\]
\[
\qquad\qquad
(1+1)^{n}
=
\sum_{r=0}^{n}
C^{n}_{r}\,1^{\,n-r} 1^{\,r}
\]
\[
\qquad\qquad
=
\sum_{r=0}^{n}
C^{n}_{r}
\]
\[
\therefore\;
2^{n}
=
\sum_{r=0}^{n}
C^{n}_{r}
\]
The binomial theorem allows a specific term to be found from the general form.
\[
(a+b)^{n} \text{ has general term }
C^{n}_{r}\,a^{\,n-r} b^{\,r}
\]
\[
\text{To find any specific term, set } r.
\]
Example
\[
\text{When } r = 3
\]
\[
C^{n}_{3}\,a^{\,n-3} b^{3}
=
\frac{n!}{3!\,(n-3)!}\,a^{\,n-3} b^{3}
\]
\[
=
\frac{n(n-1)(n-2)}{3!}\,a^{\,n-3} b^{3}
\]
\[
\text{Which is the fourth term of the expansion!}
\]
\[
\quad
C^{n}_{r}\,a^{\,n-r} b^{\,r}
\;\text{is the }(r+1)\text{th term}
\]
\[
\text{in the expansion for } 0 \le r \le n
\]
Example
Find the 7th term of the expansion of \( (2x+3y)^9 \).
\[
(2x+3y)^{9} \text{ has general term }
C^{n}_{r}\,a^{\,n-r}b^{\,r}
\]
\[
a = 2x,\quad b = 3y,\quad n = 9
\]
\[
r+1 = 7 \;\Rightarrow\; r = 6
\]
\[
C^{9}_{6}\,(2x)^{\,9-6}\,(3y)^{6}
=
\frac{9!}{6!\,3!}\,(2x)^{3}(3y)^{6}
\]
\[
= 84 \cdot 2^{3} \cdot 3^{6} \cdot x^{3} y^{6}
\]
\[
= 489888\,x^{3}y^{6}
\]
Example
Find the term containing x3 in the expansion of (3 +2x)5
\[
(3 + 2x)^{5} \text{ has general term }
C^{n}_{r}\,a^{\,n-r} b^{\,r}
\]
\[
a = 3,\quad b = 2x,\quad n = 5
\]
\[
\text{want term containing } x^{3},\; \Rightarrow\; r = 3
\]
\[
C^{5}_{3}\,3^{\,5-3}\,(2x)^{3}
=
\frac{5!}{3!\,2!}\,3^{2}\,2^{3}\,x^{3}
\]
\[
= 10 \times 9 \times 8 \, x^{3}
\]
\[
= 720\,x^{3}
\]
Probability and the Binomial Theorem
\[
\text{Let } p \text{ denote the probability of success of an outcome and } q \text{ denote the probability of failure}
\]
\[
p + q = 1
\]
\[
\text{The probability of obtaining } r \text{ successes in } n
\text{ independent trials is}
\]
\[
P(r \text{ successes})
= \; C^{n}_{r}\, q^{\,n-r}\, p^{\,r}
\]
Example
\[
\text{The probability that it will rain on any given day is } 0.4.
\]
\[
\text{What is the probability that it will rain twice in a seven‑day week?}
\]
\[
p = 0.4
\qquad
q = 0.6
\]
\[
n = 7 \; \text{(number of possible outcomes)}
\qquad
r = 2 \; \text{(number of required outcomes)}
\]
\[
P(\text{rain twice})
= C^{\,n}_{\,r}\, q^{\,n-r}\, p^{\,r}
\]
\[
= C^{\,7}_{\,2}\,(0.6)^{5}\,(0.4)^{2}
\]
\[
= \frac{7!}{2!\,5!}\,(0.6)^{5}\,(0.4)^{2}
\]
\[
= 21 \times 0.07776 \times 0.16
\]
\[
= 0.26127
\]
\[
= 26.1\%
\]
The Binomial Expansion and e
\[
\left(1 + \frac{1}{n}\right)^{n}
= 1
+ n\!\left(\frac{1}{n}\right)
+ \frac{n(n-1)}{2!}\!\left(\frac{1}{n}\right)^{2}
+ \cdots
+ \left(\frac{1}{n}\right)^{n}
\]
\[
\lim_{\,n \to \infty}
\left(\frac{1}{n}\right)^{n}
= 0
\]
\[
\left(1 + \frac{1}{n}\right)^{n}
\;=\;
\sum_{r=0}^{\infty} \frac{1}{r!}
\]
\[
\sum_{r=0}^{\infty} \frac{1}{r!} = e
\]
\[
e^{x}
=
\sum_{r=0}^{\infty}
\frac{x^{\,r}}{r!}
\]
\[
e^{x}
=
1
+ x
+ \frac{x^{2}}{2!}
+ \frac{x^{3}}{3!}
+ \ldots\ldots\ldots
\]
Example
\[
e^{3}
=
1
+ 3
+ \frac{3^{2}}{2!}
+ \frac{3^{3}}{3!}
+ \frac{3^{4}}{4!}
+ \frac{3^{5}}{5!}
+ \frac{3^{6}}{6!}
+ \frac{3^{7}}{7!}
+ \ldots
\]
\[
e^{3}
=
1
+ 3
+ \frac{9}{2}
+ \frac{27}{6}
+ \frac{81}{24}
+ \frac{243}{120}
+ \frac{729}{720}
+ \frac{2187}{5040}
+ \ldots
\]
\[
e^{3}
=
1
+ 3
+ 4.5
+ 4.5
+ 3.375
+ 2.025
+ 1.0125
+ 0.4339\ldots
\]
