Factorial

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Notes

Though not stated explicitly as a learning objective in chapter 3, knowledge of factorials are required in a few places. This page should deal with some basics of factorials and won't deal with its usages in probability.

Definition

n! \, =n\left(n-1 \right)\left(n-2 \right)\left(n-3 \right)\times \ldots \times 3\times 2\times 1

where n \, is a positive integer /natural number

  • n! \, is stated as n \, factorial.
  • by definition, 0!=1 \, (you can confirm this using calculator, and need not worry why)
  • 1!\, =1 \,
  • 2!\, =2\times1=2 We might as well remember 2!=2\,

Examples

4! \, =4\times3\times2\times1

10! \, =10\times9\times8\times\ldots\times3\times2\times1

  • Note : Notice the usage of \ldots to represent the recurring pattern
    • It's always three dots, never more or less
    • Note that it's written as {\color{Red}\times}{\color{Blue}\ldots}{\color{Red}\times}, DON'T write it as =10\times9\times8\ldots3\times2\times1

Now try writing out the definition of 13! \, =13\times12\times11\times\ldots\times3\times2\times1

Finding the value

To find the value of the factorial, we just need to use our handy calculator, of course. (Use the x! \, function)

  • 4! \, =24 \,
  • 10! \, =3628800 \,
  • 13! \, =6227020800 \,

Examples involving n

If we can find the values using calculators, why bother with the definition in the first place? Because it allows us to apply it in more general situations and also allow us to manipulate it. For example, what would \left(2n+1 \right)! mean? Let's see...

The definition is

n!=n\left(n-1 \right)\left(n-2 \right)\left(n-3 \right)\times \ldots \times 3\times 2\times 1

However, when we write out 10! \, we DON'T actually write

\left(10 \right)\left(10-1 \right)\left(10-2 \right)\times \ldots \times 3\times2\times1

Thats because we understand the formula as a series of steps (rather than just something for us to substitute)

  • take the integer
  • multiply with the next (smaller) integer, and the next, and the next
  • until we reach 1, (normally \times3\times2\times1)

Thus, very naturally, we will write 10!=10\times9\times8\times\ldots\times3\times2\times1

This should be no different when we write out other factorials.

So let's try \left(n+3 \right)!

  • First we take the integer itself, in this case \left(n+3 \right)
  • Multiply with the next smaller integer, which will be \left(n+2 \right)
  • and the next, which will be \left(n+1 \right)
  • until we reach \times3\times2\times1

Thus \left(n+3 \right)! =\left(n+3 \right)\left(n+2 \right)\left(n+1 \right)n\times\ldots\times3\times2\times1


Now, try \left(n-2 \right)!

Answer =\left(n-2 \right)\left(n-3 \right)\left(n-4 \right)\times\ldots\times3\times2\times1

Note : The next smaller integer after \left(n-2 \right) is \left(n-3 \right), NOT \left(n-1 \right)


How about \left(2n \right)! and 2n! \,

  • \left(2n \right)! =\left(2n\right)\left(2n-1 \right)\left(2n-2 \right)\times\ldots\times3\times2\times1 it's just really {\color {Red} \heartsuit}!={\color {Red} \heartsuit}\left({\color {Red} \heartsuit}-1 \right)\left({\color {Red} \heartsuit}-2 \right)\left({\color {Red} \heartsuit}-3 \right)\times \ldots \times 3\times 2\times 1

But

  • 2n! \, =2\left[n\left(n-1 \right)\left(n-2 \right)\left(n-3 \right)\times \ldots \times 3\times 2\times 1 \right] the factorial only applies for the n \,, not the 2 \,, thus we leave the 2 \, outside


So, returning to our first question, \left(2n+1 \right)! =\left(2n+1\right)\left(2n\right)\left(2n-1 \right)\times\ldots\times3\times2\times1

Manipulating Factorial Terms

Depending on the situation, we can combine/simplify/factorize/divide/etc factorial terms with each other or with other suitable numbers/terms.

Lets say we have

  • 4\left(3! \right)
  • But we know that 3!=3\times2\times1
  • Thus 4\left(3! \right) =4\times3\times2\times1
  • Hmmm.... which means it is equal to 4! \,
  • So we have 4\left(3! \right)=4!

Now, try 11\left(10! \right)

  • Answer : 11\left(10! \right)=11!
  • Working is not needed. However, if you have trouble imagining it, you can write it out 11\left(10! \right)=11\times \left( 10 \times 9 \times 8 \ldots \times 3 \times 2 \times 1\right)=11!

So when will this work? Does it work for 8\left(6! \right)?

No. It will only work when the number we multiply in front is larger by one than the integer of the factorial. And it will result in a larger factorial.


For n! \,, if we want to make the factorial larger, then we multiply in front, a larger by one integer, that is \left(n+1\right)

  • Thus, we have \left(n+1\right)n! , which will be equals to \left(n+1\right)!
  • \left(n+1\right)n!=\left(n+1\right)! might look weird initially, but it really isn't any different from 4\left(3! \right)=4!
  • We can also see it from the definition \left(n+1\right){\color{Red}n!}=\left(n+1\right){\color{Red} n\left(n-1 \right)\times \ldots \times3 \times2 \times1}=\left(n+1\right)!


Other times we would want to make the factorial smaller rather than larger. Say we have 8! \,. And we want to change the factorial to a 7! \,.

  • 8! \, =8\times 7\times 6\times \dots \times 3 \times 2 \times 1
  • But we want only 7! \, , thus we put a bracket around 7\times 6\times \dots \times 3 \times 2 \times 1
  • Thus, 8! \, =8\times {\color{Red} \left( 7\times 6\times \dots \times 3 \times 2 \times 1 \right)} =8{\color{Red} \left( 7!\right)}

So what we are doing is in effect taking out the largest integer from the factorial, which will result in a smaller factorial.

If we need it to have 6! \, instead, it will be 8! \, =8 \times 7 \times 6!


For n! \,, if we want to make the factorial smaller, then we take out, the largest integer, that is n \,

  • Thus, we know we will have something in the form of n!=n\left( \dots \right), but what exactly is left there?
  • \left(n-1\right)!. Thus n!=n\left( n-1 \right)!
  • Compare with 8! = 8 \times 7!
  • We can also see it from the definition n!=n{\color{Red} \left(n-1 \right)\times \ldots \times3 \times2 \times1}=n{\color{Red}\left( n-1 \right)!}


More examples

By adding suitable numbers in front of either side, make the equality hold for each of the following

  • \begin{array}{lcr} 2! & = & 3! \\ 2! & = & 4! \\ 5! & = & 4! \\ 7! & = & 5! \end{array}
  • Answer
    • {\color{Red}3}\left(2!\right) = 3!
    • {\color{Red}4 \times 3 \times} 2! = 4!
    • 5! = {\color{Red}5}\left(4!\right)
    • 7! = {\color{Red}7 \times 6 \times} 5!

Write out the following factorials

  • n! \, =n\left(n-1 \right)\times \ldots \times3 \times2 \times1
  • \left(n+1 \right)! =\left(n+1 \right)n\left(n-1 \right)\times \ldots \times3 \times2 \times1
  • \left(n-1 \right)! =\left(n-1 \right)\left(n-2 \right)\left(n-3 \right)\times \ldots \times3 \times2 \times1

By adding suitable terms in front of either side, make the equality hold for each of the following. (The above might help if you have trouble with it.)

  • \begin{array}{rcl} n! & = & \left(n+1 \right)! \\ n! & = & \left(n-1 \right)! \\ \left(n+2 \right)! & = & n! \\ \left(n-1 \right)! & = & \left(n+1 \right)! \end{array}
  • Answer
    • {\color{Red} \left(n+1 \right)}n! =\left(n+1 \right)! \left(n+1 \right)! is the larger factorial
    • n! ={\color{Red}n}\left(n-1 \right)! n! \, is the larger factorial
    • \left(n+2 \right)! ={\color{Red}\left(n+2 \right)\left(n+1 \right)}n! Compare with 8!=8 \times 7 \times 6!
    • {\color{Red} \left(n+1 \right)n}\left(n-1\right)! =\left(n+1 \right)!

Adding/Subtracting Factorial Terms

Equal factorial terms, can be added/subtracted just like any other terms. Thus

  • 6!+3\left(6!\right) =4\left(6!\right) It really is just {\color{Red} \heartsuit }+3{\color{Red} \heartsuit }=4{\color{Red} \heartsuit }
  • n!+3n! \, =4n! \,
  • n!+\left(n+1\right)n! =\left(n+2\right)n! {\color{Red}n!}+\left(n+1\right){\color{Red}n!} = \left(1+n+1\right){\color{Red}n!}


When they are not equal (but does not differ too much from each other), we can make them equal by using methods learnt earlier.

Say we have 6!+7! \, To change the 6!\, to 7! \,, we would need to multiply an additional number, which of course we can't simply do it here. So instead, we change the 7! \, to 6! \,.

Thus, 6!+7! \, =6!+7\left(6!\right) =8\left(6!\right)

In other words, we change the factorials into the smallest common factorial before adding/subtracting.

More examples

  • 9!-8! \, =8\left(8!\right)
    • \begin{align} 9!-8! & = 9\left(8!\right)-8!\\ & = 8\left(8!\right)\end{align}
  • 8!+6! \, = 57\left(6!\right)
    • \begin{align} 8!+6! &= 8\times 7 \times 6!+6!\\ &=57\left(6!\right)\end{align}
  • 5!+4!-3! \, =23\left(3!\right)
    • \begin{align} 5!+4!-3! &= 5 \times 4 \times 3!+ 4\times 3! -3!\\ &= 23\left(3!\right)\end{align}
  • \left(n+1\right)!-n! =n\left(n!\right)
    • \begin{align} \left(n+1\right)!-n! &= \left(n+1\right)n!-n!\\ & = \left(n+1-1\right)n!\\ & =n\left(n!\right)\end{align}
  • \left(n+1\right)!+\left(n+2\right)! =\left(n+3\right)\left(n+1\right)!
    • \begin{align} \left(n+1\right)!+\left(n+2\right)! &= \left(n+1\right)!+\left(n+2\right)\left(n+1\right)!\\ &= \left(n+3\right)\left(n+1\right)!\end{align}
  • \left(n+2\right)!+n! =\left(n^2+3n+3\right)n!
    • \begin{align} \left(n+2\right)!+n! &= \left(n+2\right)\left(n+1\right)n!+n!\\ &= \left[\left(n^2+3n+2\right)+1\right]n! \\ & =\left(n^2+3n+3\right)n!\end{align}
  • n!-\left(n-2\right)! =\left(n^2-n-1\right)\left(n-2\right)!
    • \begin{align} n!-\left(n-2\right)!&= n\left(n-1\right)\left(n-2\right)!-\left(n-2\right)!\\ &=\left[\left(n^2-n\right)-1\right]n!\\ &=\left(n^2-n-1\right)\left(n-2\right)!\end{align}

Dividing Factorial Terms

To see how we would divide (or at least simplify fractions) factorial terms, its useful to look back at how we would do it for normal integers

For example, when we write

  • \frac{12}{4}=3, what we are effectively doing is \frac{12}{4}=\frac{3 \times {\color{Red}4}}{{\color{Red}4}}=3, that is cutting away the 4
  • \frac{4}{20} =\frac{{\color{Red}4}}{5 \times {\color{Red}4}}=\frac{1}{5}

We are of course taking away the greatest common factor.

So how would this be applied with factorial terms? Notice that if we have

  • \frac{5\left(4! \right)}{4!} It just becomes 5 \,
  • \frac{\left(n-1\right)!}{n\left(n-1\right)!} =\frac{1}{n}

Which means that if we have different (but not too far) factorial terms in the numerator and the denominator, we make both the same so that we can cut. We would of course, need to change the larger factorial into the smaller factorial.

So suppose we have

  • \frac{9!}{8!}
  • we know we can change 9! \, to 8! \, easily, so that we can simplify
  • \frac{9\left(8! \right)}{8!}=9

Try

  • \frac{n!}{\left(n+1 \right)!} Answer : =\frac{1}{n+1}
  • Which one is the larger factorial? \left(n+1 \right)!
    • \frac{n!}{\left(n+1 \right)!}=\frac{n!}{\left(n+1 \right)n!}=\frac{1}{n+1}

More examples

  • \frac{7!}{8!} =\frac{1}{8}
    • \frac{7!}{8!}=\frac{7!}{8\left( 7!\right)}=\frac{1}{8}
  • \frac{9!}{7!} =72\,
    • \frac{9!}{7!}=\frac{9 \times 8 \times \left( 7!\right)}{7!}=72
  • \frac{13!}{10!} =1716\,
    • \frac{13!}{10!}=\frac{13 \times 12 \times 11 \times \left(10!\right)}{10!}=1716
  • \frac{\left(n+1 \right)!}{n!} =n+1\,
    • \frac{\left(n+1 \right)!}{n!}=\frac{\left(n+1 \right)n!}{n!}=n+1\,
  • \frac{\left(n-1 \right)!}{n!} =\frac{1}{n}
    • \frac{\left(n-1 \right)!}{n\left(n-1 \right)!}=\frac{1}{n}
  • \frac{\left(n-4 \right)!}{\left(n-3 \right)!} =\frac{1}{n-3}
    • \frac{\left(n-4 \right)!}{\left(n-3 \right)\left(n-4 \right)!}=\frac{1}{n-3}
  • \frac{\left(n+1\right)!}{\left(n-1\right)!} =n\left(n+1\right)
    • \frac{\left(n+1\right)!}{\left(n-1\right)!}=\frac{\left(n+1\right)n\left(n-1\right)!}{\left( n-1\right)!}=n\left(n+1\right)

Adding/Subtracting Fractions with Factorial Denominators

Again, its useful to look back at how we would do it for normal integers

  • \frac{1}{3}-\frac{1}{4} =\frac{{\color{Red}4}-{\color{Blue}3}}{{\color{Blue}3}\times{\color{Red} 4}}
  • However, if it is \frac{1}{2}-\frac{1}{6} we would of course do it different, because we know that 2 \, is factor of 6 \,
  • \frac{1}{2}-\frac{1}{6} \frac{1 \times {\color{Red}3}}{2 \times {\color{Red}3}}-\frac{1}{6} =\frac{3-1}{6}

We are, of course, taking the least common multiple to be the denominator.

So how would this be applied in with factorial terms?

Lets try

  • \frac{1}{4!}-\frac{1}{5!}
  • We know that we can change 5! \, to 4! \, and vice versa, in this case we change the smaller into larger
  • But to change 4! \, to 5! \, we would need to multiply with a 5 \,. Thus, we multiply 5 \, on the numerator AND denominator.
  • \frac{1}{4!}-\frac{1}{5!} =\frac{{\color{Red}5} \times 1}{{\color{Red}5} \times 4!}-\frac{1}{5!} =\frac{5}{5!}-\frac{1}{5!}=\frac{4}{5!}

More examples

Note : We can actually use the calculator to confirm our answers (check that both sides have same value)

  • \frac{1}{6!}+\frac{1}{7!} =\frac{8}{7!}
    • \begin{align} \frac{1}{6!}+\frac{1}{7!}& =\frac{7}{7\times6!}+\frac{1}{7!}\\ &=\frac{7}{7!}+\frac{1}{7!}\\ &=\frac{8}{7!}\end{align}
  • \frac{2}{3!}+\frac{7}{4!} =\frac{15}{4!}
    • \begin{align} \frac{2}{3!}+\frac{7}{4!}&=\frac{4\times2}{4\times3!}+\frac{7}{4!}\\ &=\frac{8}{4!}+\frac{7}{4!}\\ &=\frac{15}{4!}\end{align}
  • \frac{1}{4!}-\frac{1}{6!} =\frac{29}{6!}
    • \begin{align}\frac{1}{4!}-\frac{1}{6!}&=\frac{6\times5\times1}{6\times5\times4!}-\frac{1}{6!}\\ &=\frac{30}{6!}-\frac{1}{6!}\\ &=\frac{29}{6!}\end{align}
  • \frac{1}{n!}+\frac{1}{\left(n+1\right)!} =\frac{n+2}{\left(n+1\right)!}
    • \begin{align} \frac{1}{n!}+\frac{1}{\left(n+1\right)!}& =\frac{n+1}{\left(n+1\right)n!}+\frac{1}{\left(n+1\right)!}\\ & = \frac{n+1}{\left(n+1\right)!}+\frac{1}{\left(n+1\right)!}\\ &=\frac{n+2}{\left(n+1\right)!}\end{align}
  • \frac{1}{\left(n-1\right)!}-\frac{1}{n!} =\frac{n-1}{n!}
    • \begin{align}\frac{1}{\left(n-1\right)!}-\frac{1}{n!}&=\frac{n}{n\left(n-1\right)!}-\frac{1}{n!}\\ & =\frac{n}{n!}-\frac{1}{n!}\\ &=\frac{n-1}{n!}\end{align}
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