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.

\(\begin{align} 7! &= 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \\ &= 5040 \end{align}\)

The maximum number of ways the letters in 'forgive' can be rearranged is 5040, with no repeats.

Evaluate \(3! +2!\)

Answer:

\(3! = 3 \cdot 2 \cdot 1\)

\(2! = 2 \cdot 1\)

\(3! + 2! = 3 \cdot 2 \cdot 1 + 2 \cdot 1 = 8\)

Evaluate \(5! - 3!\)

Answer:

\(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\)

\(3! = 3 \cdot 2 \cdot 1\)

\(5! - 3! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 - 3 \cdot 2 \cdot 1= 114\)

Evaluate \(3! \cdot 4!\)

Answer:

In this expression \(3! \cdot 4!\), means the two factorials are multiplying each other.

\(3! \cdot 4!\)

\((3!)(4!) = (3 \cdot 2 \cdot 1)(4 \cdot 3 \cdot 2 \cdot 1)\)

\(6 \cdot 24 = 144\)

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

\(4 \cdot 1 = 4\)

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 \(\frac{1}{8}\)