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# 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

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.

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