A Combinations Problem Involving Days Of the Week 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?
 A: 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.
A: Of course for small cases you can make do with your notation. But instead, consider the case where you have to select all your working days for the next year of, say, $40$ weeks. Then the problem comes down to selecting $80$ out of $200$ options. Then the formula becomes 
$$\frac{200\cdot199\cdot198\cdot\ldots\cdot121}{80\cdot79\cdot78\cdot\ldots\cdot1}$$
Instead, in this case the notation of the binomial coefficient lets you write this expression in a shorter form:
$$\frac{200\cdot199\cdot198\cdot\ldots\cdot121}{80\cdot79\cdot78\cdot\ldots\cdot1} = \frac{200\cdot199\cdot198\cdot\ldots\cdot121}{80\cdot79\cdot78\cdot\ldots\cdot1} \cdot \frac{120\cdot119\cdot118\cdot\ldots\cdot1}{120\cdot119\cdot118\cdot\ldots\cdot1} = \frac{200!}{80!\cdot120!} = \binom{200}{80}$$
