Categories
Algebra Arithmetic Combinatorics Math Olympiad PRMO

Rearrangement Problem | PRMO 2019 | Question 27

Try this beautiful problem from the Pre-RMO, 2019 based on the Diameter of a circle. You may use sequential hints to solve the problem.

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


Arrangement

Sets

Integer

Check the Answer


But try the problem first…

Answer: is 84.

Source
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)=n(Q)=53

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

Final Step

So, required arrangements

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

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.