# 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?

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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:

$$\frac{200\cdot199\cdot198\cdot\ldots\cdot121}{80\cdot79\cdot78\cdot\ldots\cdot1} = \frac{200\cdot199\cdot198\cdot\ldots\cdot121}{80\cdot79\cdot78\cdot\ldots\cdot1} \cdot \frac{120\cdot119\cdot118\cdot\ldots\cdot1}{120\cdot119\cdot118\cdot\ldots\cdot1} = \frac{200!}{80!\cdot120!} = \binom{200}{80}$$

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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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Of course one can always write something like $\frac{n\times(n-1)\times\cdots\times(n-r+1)}{r\times(r-1)\times\cdots\times2\ti‌​mes 1}$, but that is certainly less convenient -- and also carries a risk of mistakes because it leaves it to the reader to guess what the dots are supposed to stand for. If one writes down too few of the terms explicitly, it might be that there there are several different guesses that all match -- and there is no way to be certain that one has enough terms, save for the rather risky "I myself can't imagine more than one interpretation". – Henning Makholm Sep 12 '11 at 14:13