Here\'s a pretty hard looking (but really not too hard if done cleverly) inclusi
ID: 1948036 • Letter: H
Question
Here's a pretty hard looking (but really not too hard if done cleverly) inclusion-exclusion problem. The correct answer is 76. You job is to show the work that
leads to that answer.
How many permutations of the digits 1; 2; 3; 4; 5, have at least one digit in its
own spot? In other words, a 1 in the first spot, or a 2 in the second, etc. For
example, 35241 is OK since it has a 4 in the fourth spot, and 14235 is OK, since
it has a 1 in the first spot (and also a 5 in the fifth spot). But 31452 in no good.
Hint: Let A1 be the set of permutations that have 1 in the first spot, let A2 be the
set of permutations that 2 in the second spot, and so on.
Please show all steps
Explanation / Answer
This is a question of Dearrangements. No. of ways in which no digit takes its right place is given by - 5!( 1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!) = 44 Total no. of ways of arranging the 5 digits is 5! = 120 So total no. of req. permutations where atleast 1 digit stays at its right place is given by = 120 - 44 = 76.