Permutation formula for lock combination

Question

I know the basic permutation formula for k objects out of an n set. But what is the formula for determining the number of permutations where k is a range (1..m) ?

What are the formula for the following scenarios?

1) a keyless door lock has n distinct buttons, the combination can be any length (1..m), and in any sequence, but each button can only be used once.

2) same as 1, except any step in the sequence (1..m) can be two buttons pushed together, the constraint that a button only pressed once, still holds

gt6989b · Answer 1 · 2016-06-15T03:57:38.087

1

Let's do (1) and you can finish (2) yourself.

Pick the buttons for the combination - $\binom{n}{m}$ ways
Arrange them in any of the $m!$ orders.

You get $$\binom{n}{m} m! = \frac{n!}{(n-m)!}$$

UPDATE This takes care of any fixed $m$. To get the result for different $m$, add different possibilities: $$ \sum_{k=1}^m \frac{n!}{(n-k)!} $$

edited Jun 15 '16 at 03:57

answered Jun 14 '16 at 20:52

gt6989b

54,422

I believe you shared the formula for permutations for a fixed sample size n out of a population of m. My question is about a variable sample size (1..m) – MarkE Jun 14 '16 at 22:15
@MarkE see update – gt6989b Jun 15 '16 at 03:57
Are you sure that is correct? The answer is undefined when m=n – MarkE Jun 15 '16 at 17:37
@MarkE $0!=1$... – gt6989b Jun 15 '16 at 17:44
Doh! Wasn't thinking. – MarkE Jun 15 '16 at 23:00
Still not sure how part two of my question changes the formula. It does seem that it would increase the number of permutations because two digits pressed together are distinct from two digits pressed in sequence. – MarkE Jun 15 '16 at 23:03
@MarkE same basic summation over $k$, the trick is to count how many possibilities can you choose from. – gt6989b Jun 15 '16 at 23:21

Permutation formula for lock combination

1 Answers1