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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).

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

Arrangement

Sets

Integer

But try the problem first...

Answer: is 84.

Source

Suggested Reading

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.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

Content

[hide]

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).

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

Arrangement

Sets

Integer

But try the problem first...

Answer: is 84.

Source

Suggested Reading

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.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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