[Math] A Combinations Problem Involving Days Of the Week

binomial-coefficientscombinatoricsproblem solving

I'am reading through Engineering Math by Ken Stroud/Dexter Booth and in page 274 under Combinations. Here's the situation.

Assuming that you have a part-time Job in the weekday evenings where you have to be at work just two evenings out of the five. Let's also assume that your employer is very flexible and allows you to choose which evenings you work provided you ring him up on sunday and tell him. One possible selection could be:

Mon-Work Tue-Work

Another selection could be:

Wed-Work Fri-Work

So the possible arrangements among the five days is $$5 \cdot 4 = 20.$$

Because

There are $5$ weekdays from which to make a first selection and for each such
selection there are $4$ days left from which to make the second selection. This
gives a total of $5 \cdot 4 = 20$ possible arrangements.
However, not all arrangements are different. For example, if, on the Sunday,
you made your first choice as Friday and your second choice as Wednesday,
this would be the same arrangement as making
your first choice as Wednesday and your second choice as Friday.
So every arrangement is duplicated.
How many different arrangements are there?

$$\frac{5\cdot 4 }{ 2 }= 10.$$

And he says it's

Because each arrangement is duplicated. List them:

Mon, Tue Mon, Wed Mon, Thu Mon, Fri

Tue, Wed Tue, Thu Tue, Fri

Wed, Thu Wed, Fri

Thu, Fri

There are $10$ different ways of combining two identical items in five different
places.

The expression $$ \frac{5 \cdot 4}{2}$$ can be written in factorial form as follows.

$$\frac{5 \cdot 4}{2} = \frac{5\cdot 4}{2\cdot 1} =
\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{(3 \cdot 2 \cdot 1)(2\cdot 1)} = \frac{5!}{3!2!}.$$

He ends by saying, if your employer asked you to work $3$ evenings out of the $5$, how many
different arrangements could you select? (Give your answer in terms of
factorials.)

$$\frac{5!}{(5-3)!3!}.$$

Ok My question is Why on earth should i ever write $$\frac{5 \cdot 4}{2} = 10$$ in factorial form as it's written above with additional natural numbers and end that way?

Best Answer

Because if you want to describe a more general result, the factorials are convenient. If you need to choose $r$ days to work out of $n$ days, the number of ways to choose is the number of combinations, written as ${n \choose r}=\frac{n!}{r!(n-r)!}$. This is hard to describe without the factorials.

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