1

Consider all possible permutations of eight distinct elements $a, b, c, d, e, f, g, h$. In how many of them, will $d$ appear before $b$? Note that $d$ and $b$ may not necessarily be consecutive.

Alex_ban
  • 119

2 Answers2

2

From these 8 elements we can create 8! permutations.

There is a one to one correspondence between the permutations where $d$ appears after $b$ and the permutations where $d$ appears before $b$. This is a simple switching of the elements $d$ and $b$.

Hence half of the 8! permutations will be so that $d$ appears before $b$. This equals 20160.

TCiur
  • 498
  • those kinds of shortcuts always miss their road to my intuition, nice trick ! – Abr001am Feb 24 '16 at 09:45
  • I'm baffled. Can you assist pls? 1. How do you know anything about the "correspondence between the permutations where d appears after b and the permutations where d appears before b"? How do you know this is "one to one"? 2. Then how do you deduce "half of the 8! permutations"? –  Jan 03 '22 at 06:58
  • Please edit your answer, rather than reply in comments that can get deleted. –  Jan 03 '22 at 06:58
0
  • By inventing a function f(n) where n is the range of b

$f(n)=(n-1)!*\binom{8-2}{8-n}(8-n)!$ the possible permutations where b is at the $n$th rank.

This gives outcome to f(2)+f(3)+..f(8) all valid permutations .

Abr001am
  • 746