Try this beautiful problem from the Pre-RMO, 2019 based on rearrangement.

Rearrangement Problem – PRMO 2019

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).

  • is 107
  • is 84
  • is 840
  • cannot be determined from the given information

Key Concepts




Check the Answer

But try the problem first…

Answer: is 84.

Suggested Reading

PRMO, 2019, Question 27

Principles and Techniques in Combinatorics by Chi Chuan

Try with Hints

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 \(A_2\)


\(n(P \cap Q)=9\)

Final Step

So, required arrangements

=\(\frac{1}{2}[265-n(P \cup Q)]\)=84.

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