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I am following "A first course in Discrete Mathematics" by Ian Anderson. It stated the following theorem without providing any proof, which might be because it is too easy. But I am having a tough time to come up with the said proof.

Theorem 1.2 The number of ways of selecting $r$ objects from $n$, in order with no repetition, is $$n(n-1)...(n-r+1)=\frac{n!}{(n-r)!}$$

I first thought of proving this using induction but clearly if it holds for some $r$ it need not hold for $r+1$. I can't seem to think of any method in my arsenal that can tackle this. Any helps will be appreciated.

Shyam Tripathi
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1 Answers1

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Induction should do the job.

When $k = 1$, we have $n$ object to choose from and thus there are $n$ choices where $n = \frac{n!}{(n-1)!}$.

Now assume the statement holds for $k = r-1$, i.e. to choose $r-1$ objects from $n$ in order with no repetition, there are $\frac{n!}{(n-(r-1))!}$ ways.

When $k = r$, aside from the previous ways from the last step, we have to choose $1$ more object from the remaining $n - (r-1) = n-r+1$ items, notice that all $\frac{n!}{(n-(r-1))!}$ ways for selecting $r-1$ objects are all independent of each other, thus selecting $1$ more and append it to the list will add a new way to selecting $r$ items.

Therefore selecting $r$ objects from $n$ objects in order with no repetition has $(n-r+1)\frac{n!}{(n-(r-1))!} = \frac{n!}{(n-r)!}$ ways.

Mq Hu
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