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AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Row of Pascal Triangle | AIME I, 1992 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Row of Pascal Triangle.

Row of Pascal Triangle – AIME I, 1992


In Pascal’s Triangle, each entry is the sum of the two entries above it. Find the row of Pascal’s triangle do three consecutive entries occur that are in the ratio 3:4:5.

  • is 107
  • is 62
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Combinatorics

Check the Answer


Answer: is 62.

AIME I, 1992, Question 4

Elementary Number Theory by David Burton

Try with Hints


For consecutive entries

\(\frac{{n \choose (x-1)}}{3}=\frac{{n \choose x}}{4}=\frac{{n \choose {x+1}}}{5}\)

from first two terms \(\frac{n!}{3(x-1)!(n-x+1)!}=\frac{n!}{4x!(n-x)!}\)

\(\Rightarrow \frac{1}{3(n-x+1)}=\frac{1}{4x}\)

\(\Rightarrow \frac{3(n+1)}{7}=x\) is first equation

for the next two terms

\(\frac{n!}{4(n-x)!x!}=\frac{n!}{5(n-x-1)!(x+1)!}\)

\(\Rightarrow \frac{4(n-x)}{5}=x+1\)

\(\Rightarrow \frac{4n}{5}=\frac{9x}{5}+1\)

from first equation putting value of x here gives

\(\Rightarrow \frac{4n}{5}=\frac{9 \times 3(n+1)}{5 \times 7}+1\)

\(\Rightarrow n=62, x=\frac{3(62+1)}{7}=27\)

\(\Rightarrow\) n=62.

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Algebra Arithmetic Combinatorics Math Olympiad Math Olympiad Videos USA Math Olympiad

Probability in Games | AIME I, 1999 | Question 13

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.

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Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Probability Problem | Combinatorics | AIME I, 2015 – Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Probability.

Probability Problem – AIME I, 2015


In a drawer Sandy has 5 pairs of socks, each pair a different color. on monday sandy selects two individual socks at random from the 10 socks in the drawer. On tuesday Sandy selects 2 of the remaining 8 socks at random and on wednesday two of the remaining 6 socks at random. The probability that wednesday is the first day Sandy selects matching socks is \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 341
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Theory of Equations

Probability

Check the Answer


Answer: is 341.

AIME, 2015, Question 5

Geometry Revisited by Coxeter

Try with Hints


Wednesday case – with restriction , select the pair on wednesday in \(5 \choose 1 \) ways

Tuesday case – four pair of socks out of which a pair on tuesday where a pair is not allowed where 4 pairs are left,the number of ways in which this can be done is \(8 \choose 2\) – 4

Monday case – a total of 6 socks and a pair not picked \(6 \choose 2\) -2

by multiplication and principle of combinatorics \(\frac{(5)({5\choose 2} -4)({6 \choose 2}-2)}{{10 \choose 2}{8 \choose 2}{6 \choose 2}}\)=\(\frac{26}{315}\). That is 341.

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Algebra Arithmetic Combinatorics Math Olympiad Math Olympiad Videos USA Math Olympiad

Probability of tossing a coin | AIME I, 2009 | Question 3

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.

Probability of tossing a coin – AIME I, 2009 Question 3


A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to \(\frac{1}{25}\) the probability of five heads and three tails. Let p=\(\frac{m}{n}\) where m and n are relatively prime positive integers. Find m+n.

  • 10
  • 20
  • 30
  • 11

Key Concepts


Probability

Theory of equations

Polynomials

Check the Answer


Answer: 11.

AIME, 2009

Course in Probability Theory by Kai Lai Chung .

Try with Hints


here \(\frac{8!}{3!5!}p^{3}(1-p)^{5}\)=\(\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}\)

then \((1-p)^{2}\)=\(\frac{1}{25}p^{2}\) then 1-p=\(\frac{1}{5}p\)

then p=\(\frac{5}{6}\) then m+n=11

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AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Probability of divisors | AIME I, 2010 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability of divisors.

Probability of divisors – AIME I, 2010


Ramesh lists all the positive divisors of \(2010^{2}\), she then randomly selects two distinct divisors from this list. Let p be the probability that exactly one of the selected divisors is a perfect square. The probability p can be expressed in the form \(\frac{m}{n}\), where m and n are relatively prime positive integers. Find m+n.

  • is 107
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Probability

Number Theory

Check the Answer


Answer: is 107.

AIME I, 2010, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints


\(2010^{2}=2^{2}3^{2}5^{2}67^{2}\)

\((2+1)^{4}\) divisors, \(2^{4}\) are squares

probability is \(\frac{2.2^{4}.(3^{4}-2^{4})}{3^{4}(3^{4}-1)}=\frac{26}{81}\) implies m+n=107

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AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Arrangement of digits | AIME I, 2012 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement of digits.

Arrangement of digits – AIME 2012


Let B be the set of all binary integers that can be written using exactly 5 zeros and 8 ones where leading zeros are allowed. If all possible subtractions are performed in which one element of B is subtracted from another, find the number of times the answer 1 is obtained.

  • is 107
  • is 330
  • is 840
  • cannot be determined from the given information

Key Concepts


Arrangements

Algebra

Number Theory

Check the Answer


Answer: is 330.

AIME, 2012, Question 5

Combinatorics by Brualdi

Try with Hints


When 1 subtracts from a number, the number of digits remain constant when the initial number has units and tens place in 10

Then for subtraction from B requires one number with unit and tens place 10.

10 there, remaining 1 distribute any of other 11 then answer \({11 \choose 7} = {330}\).

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AMC 8 Combinatorics Math Olympiad Probability

Probability Problem | AMC 8, 2016 | Problem no. 21

Try this beautiful problem from Probability.

Problem based on Probability | AMC-8, 2016 | Problem 21


A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

  • \(\frac{3}{5}\)
  • \(\frac{2}{5}\)
  • \(\frac{1}{4}\)

Key Concepts


probability

combination

fraction

Check the Answer


Answer: \(\frac{2}{5}\)

AMC-8, 2016 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints


There are 5 Chips, 3 red and 2 green

Can you now finish the problem ……….

We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement

Can you finish the problem……..

There are 5 Chips, 3 red and 2 green

we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn.

if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green

but we draw randomly. there are 3 red boxes and 2 green boxes

Therefore the probability that the 3 reds are drawn=\(\frac{2}{5}\)

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AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Arrangement Problem | AIME I, 2012 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement.

Arrangement – AIME 2012


Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

  • is 107
  • is 216
  • is 840
  • cannot be determined from the given information

Key Concepts


Arrangements

Algebra

Number Theory

Check the Answer


Answer: is 216.

AIME, 2012, Question 3

Combinatorics by Brualdi

Try with Hints


Here the number of ways to order the string BBBCCCFFF, such that one B is in first three positions

one C is in 4th to 6th positions, and one F is for last three positions. There are (3)(3)(3)=27 ways for first 3. Then for next two, 2 ways.

Then \((3.2)^3={216}\)ways

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AMC 8 Combinatorics Math Olympiad Probability

Problem from Probability | AMC 8, 2004 | Problem no. 21

Try this beautiful problem from Probability from AMC 8, 2004.

Problem from Probability | AMC-8, 2004 | Problem 21


Spinners A and B  are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners’ numbers is even?

Problem from Probability

  • \(\frac{2}{3}\)
  • \(\frac{1}{3}\)
  • \(\frac{1}{4}\)

Key Concepts


probability

Equilly likely

Number counting

Check the Answer


Answer: \(\frac{2}{3}\)

AMC-8, 2004 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints


Even number comes from multiplying an even and even, even and odd, or odd and even

Can you now finish the problem ……….

A odd number only comes from multiplying an odd and odd…………..

can you finish the problem……..

We know that even number comes from multiplying an even and even, even and odd, or odd and even

and also a odd number only comes from multiplying an odd and odd,

There are few cases to find the probability of spinning two odd numbers from  1

Multiply the independent probabilities of each spinner getting an odd number together and subtract it from  1 we get…….

\(1 – \frac{2}{4} \times \frac{2}{3}\)= \(1 – \frac{1}{3} = \frac{2}{3} \)  

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AMC 8 Combinatorics Math Olympiad Probability USA Math Olympiad

Probability | AMC 8, 2004 | Problem no. 22

Try this beautiful problem from Probability .You may use sequential hints to solve the problem.

Probability | AMC-8, 2004 |Problem 22


At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is\(\frac{2}{5} \). What fraction of the people in the room are married men?

  • \(\frac{3}{8}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{4}\)

Key Concepts


probability

combination

Number counting

Check the Answer


Answer: \(\frac{3}{8}\)

AMC-8, 2007 problem 24

Challenges and Thrills in Pre College Mathematics

Try with Hints


Find the married men in the room …

Can you now finish the problem ……….

Find the total people

can you finish the problem……..

Assume that there are 10 women in the room, of which \(10 \times \frac{2}{5}\)=4 are single and 10-4=6 are married. Each married woman came with her husband,

so there are 6 married men in the room

Total man=10+6=16 people

Now The fraction of the people that are married men is \(\frac{6}{16}=\frac{3}{8}\)

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