0
$\begingroup$

I need to prove, that for a key with $n$ dimensions, they order of the key would be $n!$ and to prove that, I need to use mathematical induction:

$$\text{Let n = 1} \Rightarrow {1} \text{ if there are only one element in the key, there is one way to order it. } \\ \text{Suppose that n = k }\Rightarrow \text{ the order of the key is } k! \\ \text{ Then it must be proved that for } n = k+1 \Rightarrow \text{ the order of the key is } (k+1)! \\ \text{ Is it true that } k! +1 \stackrel{?}{=} (k+1)! \\ \frac{(k+1)!}{k+1} +1 \stackrel{?}{=} (k+1)! \\ (k+1)! + k +1 \stackrel{?}{=} (k+1)(k+1)! \\ 1 \stackrel{?}{=} (k+1) -\frac{(k+1)}{(k+1)!} \\ (k+1)((k+1)!-(k+1) = (k+1)! \\ (k+1)(k+1)!(k!-1) = (k+1)! \\ (k+1)(k!-1) \stackrel{?}{=} 1 $$ And this one is not true even for $k=1 $, so I must have made some error.

$\endgroup$
  • $\begingroup$ Order of key? ${}$ $\endgroup$ – kelalaka Feb 19 at 20:51
  • $\begingroup$ 1) Do you mean the number of combinations of n elements? 2) If yes, then this is not how the mathematical induction works. $\endgroup$ – mentallurg Feb 20 at 1:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.