Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.
Probability in Games - AIME I, 1999 Question 13
Forty teams play a tournament in which every team plays every other team exactly once. No ties occur,and each team has a 50% chance of winning any game it plays.the probability that no two team win the same number of games is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find \(log_{2}n\)
10
742
30
11
Key Concepts
Probability
Theory of equations
Combinations
Check the Answer
Answer: 742.
AIME, 1999 Q13
Course in Probability Theory by Kai Lai Chung .
Try with Hints
\({40 \choose 2}\)=780 pairings with \(2^{780}\) outcomes
no two teams win the same number of games=40! required probability =\(\frac{40!}{2^{780}}\)
the number of powers of 2 in 40!=[\(\frac{40}{2}\)]+[\(\frac{40}{4}\)]+[\(\frac{40}{8}\)]+[\(\frac{40}{16}\)]+[\(\frac{40}{32}\)]=20+10+5+2+1=38 then 780-38=742.
Try this beautiful problem from the Pre-RMO, 2019 based on Rectangle and Squares.
Rectangles and squares - PRMO 2019
A \(1 \times n\) rectangle \(n \geq 1\) is divided into n unit \( 1 \times 1 \) squares. Each square of this rectangle is coloured red, blue or green. Let f(n) be the number of colourings of the rectangle in which there are an even number of red squares, find the largest prime factor of \(\frac{f(9)}{f(3)}\)
Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations
Combinations- AIME, 2009
A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to $9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.
110
420
430
111
Key Concepts
Combinations
Theory of equations
Polynomials
Check the Answer
Answer: 420.
AIME I, 2009, Problem 9
Combinatorics by Brualdi .
Try with Hints
Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35
these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.
