The concept of factorials had been known since ancient times, but it wasn't until the 17th century that Christian Kramp introduced their current notation, \(n!\)
Factorials are functions in mathematics with the symbol (!), They multiply a number by every number that precedes it. They can be expressed as \(n!\), where n is the last number of the factorial.
Factorials grow very quickly - so quickly, in fact, that even relatively small factorials can quickly become too large to calculate. For example, the factorial of 20 is over 2 million, while the factorial of 100 is a staggering \(9.3 \cdot 10^{157}\). The largest factorial that can be calculated using a standard calculator is \(69!\), due to limitations in the number of digits that can be displayed.
What is a factorial?
A factorial is simply the product of all positive integers up to a given number. For example, the factorial of 5 is \(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\), or 120.
The factorial rule says the factorial of any number is that number times the factorial of the previous number. This can be expressed in a formula as \(n! = n \cdot (n-1)!\) A special case for this is \(0! = 1 \). The symbol n is a whole number, and the exclamation mark represents the expression as factorial.
Factorials can be found in permutations and combinations.
Factorial function
To summarise the above explanation, the factorial function can be expressed as:
\[n! = n \cdot (n-1)!\]
The symbol n is a whole number, and the exclamation mark represents the expression as factorial.
How to calculate factorials
You can go through these steps to find the factorial of a number n.
- Let's look at the example of 6!
- Write down the sequence of numbers you will multiply by using the factorial formula.
\(n! = n(n-1) \cdot (n-2) \cdot (n-3) \cdot (n-4) ... (n-(n-1)\)
To expand this, you will need to substitute the number n by 6 until the last subtraction is equal to \(n-(n-1)\), in this case, 1.
\(6! = 6(6-1)(6-2)(6-3)(6-4)(6-5) = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2\cdot 1\)
- Operate your numbers now you have them written down. Subtract and multiply where you have to. With our example, it's as follows:
\(6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720\)
Examples of factorials
What possible combinations can you make with the colours blue, red, and yellow in order?
Answer: We need to find the factorial of 3 considering that there are three colours, and we are finding how many combinations there could be in order.
\(n! = n(n-1)!\)
\(\begin{align} 3! &= 3 (3-1)(3-2)\\ &= 3 \cdot 2 \cdot 1 \\ &= 6 \end{align}\)
There are six possible combinations that could be made from these three colours in order, and they are:
Blue, red, and yellow --- 1
Blue, yellow, and red --- 2
Yellow, blue, and red --- 3
Yellow, red, and blue --- 4
Red, blue, and yellow --- 5
Red, yellow, and blue --- 6
How many ways can the letters in the word 'forgive' be arranged without repeating them?
Answer: To do a problem like this you count the number of letters in the word 'forgive', and then you find the factorial of it. The number of letters here is 7.
\(n! = n(n-1)! \space 7! = 7(7-1)(7-2)(7-3)(7-4)(7-5)(7-6)\)
\(\begin{align} 7! &= 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \\ &= 5040 \end{align}\)
The maximum number of ways the letters in 'forgive' can be rearranged is 5040, with no repeats.
Evaluate \(3! +2!\)
Answer:
\(3! = 3 \cdot 2 \cdot 1\)
\(2! = 2 \cdot 1\)
\(3! + 2! = 3 \cdot 2 \cdot 1 + 2 \cdot 1 = 8\)
Subtracting factorials
Evaluate \(5! - 3!\)
Answer:
\(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\)
\(3! = 3 \cdot 2 \cdot 1\)
\(5! - 3! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 - 3 \cdot 2 \cdot 1= 114\)
Multiplying factorials
Evaluate \(3! \cdot 4!\)
Answer:
In this expression \(3! \cdot 4!\), means the two factorials are multiplying each other.
\(3! \cdot 4!\)
\((3!)(4!) = (3 \cdot 2 \cdot 1)(4 \cdot 3 \cdot 2 \cdot 1)\)
\(6 \cdot 24 = 144\)
Dividing factorials
Evaluate \(\frac{4! \space 5!}{6!}\)
Answer:
Here we have to carry out two operations. We are going to multiply and divide as we expand.
\(\frac{(4\cdot 3 \cdot 2 \cdot 1)(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\)
However, notice that \(5!\) can also be found in \(6!\). We could continue with our original method way, but we can also eliminate a chunk of the figures as early as possible. Let's take a look below.
\(\frac{(4 \cdot 3 \cdot 2 \cdot 1)\cancel{(5!)}}{6 \cdot \cancel{5!}}\)
\(5!\) on both sides cancel out.
\(\frac{4 \cdot \cancel{(3 \cdot 2)} \cdot 1}{\cancel{6}}\)
We can multiply 3 and 2 first, so in a subsequent operation, that will also cancel out
\(4 \cdot 1 = 4\)
Algebraic factorials
You may sometimes meet situations with variables and factorials. So let us try to simplify expressions in the form \(\frac{(n+1)!}{n!}\).
Now let's say n = 7. That will mean \(\frac{(7+1)!}{7!}\).
\(\frac{8!}{7!} = \frac{8 \cdot 7!}{7!}\)
\(7!\) can cancel out to give us 8.
Let's add another possible situation and draw patterns from them. Let's say n = 10 now.
That would also mean \(\frac{(10 + 1)!}{10!}\)
\(\frac{11!}{10!} = \frac{11 \cdot 10!}{10!}\)
10! Will both cancel out too, leaving us with 11.
You will notice through these examples that evaluating \(\frac{(n+1)!}{n!}\) will always leave our answer at n + 1, or whatever the numerator is.
So we can conclude as a general rule for factorials \(\frac{(n+1)!}{n!} = n +1\)
However, there is a way to solve this algebraically.
\(\frac{(n+1)!}{n!} = \frac{(n+1)(n)(n-1)!}{n(n-1)!}\)
n will cancel out n, and \((n-1)!\) will cancel out \((n-1)!\) Leaving us with \(n + 1\).
Evaluate \(\frac{(n-1)!}{n!}\)
Answer:
\(\frac{(n-1)!}{n!} = \frac{(n-1)(n-2)!}{n(n-1)(n-2)!} = \frac{1}{n}\)
The terms \((n-2)!\) will both cancel each other, and \((n-1)\) term will also cancel. Leaving us with \(\frac{1}{n}\)
We can test this by inserting a numerical value into the variable. Let's take n = 8.
\(\frac{(8-1)!}{8!} = \frac{7!}{8!}\)
\(\frac{7!}{8 \cdot 7!}\)
\(7!\) Will both cancel out leaving us with \(\frac{1}{8}\)
Factorials - key takeaways
- Factorials are functions in mathematics denoted by the symbol \(n!\) They multiply a number n by every number that precedes it.
- The factorial formula can be expressed as \(n! = n \cdot (n-1)!\)
- A particular factorial is \(0! = 1\)
- Factorials are used to find arrangements, permutations, and combinations.
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