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
But try the problem first...
Answer: is 84.
PRMO, 2019, Question 27
Principles and Techniques in Combinatorics by Chi Chuan
First hint
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
Second Hint
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
\(n(P \cap Q)=9\)
Final Step
So, required arrangements
=\(\frac{1}{2}[265-n(P \cup Q)]\)=84.
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
But try the problem first...
Answer: is 84.
PRMO, 2019, Question 27
Principles and Techniques in Combinatorics by Chi Chuan
First hint
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
Second Hint
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
\(n(P \cap Q)=9\)
Final Step
So, required arrangements
=\(\frac{1}{2}[265-n(P \cup Q)]\)=84.
