Try this beautiful problem from the Pre-RMO, 2019 based on rearrangement.
We will say that the rearrangement of the letters of a word has no fixed letters if. When the rearrangement is placed directly below the word, no column has the same letter repeated. For instance, HBRATA is a rearrangement with no fixed letters of BHARAT. How many distinguishable rearrangements with no fixed letters do BHARAT have? (The two A's are considered identical).
Arrangement
Sets
Integer
Answer: is 84.
PRMO, 2019, Question 27
Principles and Techniques in Combinatorics by Chi Chuan
Let us assume 2 A's as \(A_1\) and \(A_2\) \(BHA_1RA_2T\)
Numbers of rearrangement of these 6=6!\((\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!})\)=265
Let P be a set when \(A_2\) occupies \(A_1)\)
and Q be a set when \(A_1\) occupies \(A_2\)
n(P)=n(Q)=53
\(n(P \cap Q)=9\)
So, required arrangements
=\(\frac{1}{2}[265-n(P \cup Q)]\)=84.
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