Number of ways of arrangement | PRMO 2017 | Question 10

Try this beautiful problem from the Pre-RMO, 2017 based on Number of ways of arrangement.

Number of ways of arrangement - PRMO 2017

There are eight rooms on the first floor of a hotel, with four rooms on each side of the corridor, symmetrically situated (that is each room is exactly opposite to one other room). Four guests have to be accommodated in four of the eight rooms (that is one in each) such that no two guests are in adjacent rooms or in opposite rooms, find number of ways can the guests be accommodated.

is 107
is 48
is 840
cannot be determined from the given information

Key Concepts

Number of ways

Integers

Arrangement

Check the Answer

Answer: is 48.

PRMO, 2017, Question 10

Problem Solving Strategies by Arthur Engel

Try with Hints

here there is particular way rooms are arranged with guests

Let 1 g be guest in room 1, 3 g be guest in room 3, 6 g be guest in room 6, 8 g be guest in room 8 then arrangement = 1 g 2 empty 3 g 4 empty

5 empty 6 g 7 empty 8 g arrangement wise

where room 1 and room 5 are opposite and facing each other with room 1 has guest and room 5 empty

room 2 and room 6 are opposite and facing each other with room 2 empty and room 6 has guest

room 3 and room 7 are opposite and facing each other with room 3 has guest and room 7 empty

room 4 and room 8 are opposite and facing each other with room 4 empty and room 8 has guest

here with four guests to be filled in four rooms

which can be arranged in 4! ways

empty and filled rooms can be arranged in 2! ways

required number of ways=\(2 \times 4!\)=48 ways.

Rearrangement Problem | PRMO 2019 | Question 27

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

Answer: is 84.

PRMO, 2019, Question 27

Principles and Techniques in Combinatorics by Chi Chuan

Try with Hints

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

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

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

So, required arrangements

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

Arranging in column | AIME I, 1990 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Arranging in column.

Arranging in column - AIME I, 1990

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marks man is to break all the targets according to the following rules

1 ) The marksman first chooses a column from which a target is to be broken,

2 ) the marksman must then break the lowest remaining target in the chosen column. If the rules are followed, in how many different orders can the eight targets be broken?

is 107
is 560
is 840
cannot be determined from the given information

Key Concepts

Integers

Arrangement

Algebra

Check the Answer

Answer: is 560.

AIME I, 1990, Question 8

Combinatorics by Brualdi

Try with Hints

Let the columns be labelled A,B and C such that first three targets are A, A and A the next three being B, B and B and the next being C and C in which we consider the string AAABBBCC.

Since the arrangement of the strings is one-one correspondence and onto to the order of shooting for example first A is shot first, second A is shot second, third A is shot third, first B is shot fourth, second B is shot fifth, third B is shot sixth, first C is shot seventh, second C is shot eighth,

or, here arrangement of the strings is bijective to the order of the shots taken

the required answer is the number of ways to arrange the letters which is \(\frac{8!}{3!3!2!}\)=560.

Two Arrangements | PRMO 2019 | Question 5

Try this beautiful problem from the PRMO, 2019 based on Two arrangements.

Two arrangements - PRMO 2019

Five persons wearing badges with numbers 1,2,3,4,5 are seated on 5 chairs around a circular table. Find the number of ways they be seated that no two persons whose badges have consecutive numbers are seated next to each other.

is 107
is 10
is 840
cannot be determined from the given information

Key Concepts

Arrangement

Algebra

Number Theory

Check the Answer

Answer: is 10.

PRMO, 2019, Question 5

Combinatorics by Brualdi

Try with Hints

Here they may seat in circular way such that 1,3,5,2,4 and again 1

then they may seat in another circular way 1,4,2,5,3 and again 1

Then number of ways = (2)(5) =10.

Number of ways of arrangement | PRMO 2017 | Question 10