Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

what is the number of permutations of a set say {1, 2, 3, 4, 5} that keep at least one integer fixed ?

share|cite|improve this question
What is the number of permutations that keep the 1 fixed? Answer this and you will be able to answer your question. – Djaian Nov 2 '10 at 15:11
n! - # of perms that keep none fixed (latter is called a derangement, closed form formulae are available) – user1709 Nov 2 '10 at 15:11
Yes, listen to Djaian, but possibly instead think of those that leave 5 fixed. Doesn't really matter however. – BBischof Nov 2 '10 at 15:13
@Djaian: this is not enough by itself. @muad: permutations of a set is fine. – Qiaochu Yuan Nov 2 '10 at 15:20
@Qiaochu : you are right. I got tricked like a real beginner here. My bad, I should have thought a little bit before commenting :-) – Djaian Nov 2 '10 at 15:45
up vote 9 down vote accepted

This is equal to $n!$ minus the number of permutations which keep no elements fixed. These are known as derangements and you can find all the relevant formulas at the Wikipedia article. You can count them, for example, using inclusion-exclusion.

If you just want the number of derangements of an $n$-element set, then a nice way to compute it is to round $\frac{n!}{e}$ to the nearest integer. For example, when $n = 5$ we want to round $\frac{120}{e} \approx 44.1$, so there are $44$ derangements, hence $120 - 44 = 76$ permutations which fix at least one element.

share|cite|improve this answer
+1,nice explanation, also this link gives a nice idea for computing derangement… – Quixotic Nov 3 '10 at 12:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.