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AMC 10 Geometry Math Olympiad USA Math Olympiad

Chosing Program | AMC 10A, 2013 | Problem 7

Try this beautiful problem from Combinatorics: Chosing Program

Chosing Program – AMC-10A, 2013- Problem 7


A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?

,

  • $6$
  • $8$
  • $9$
  • $12$
  • \(16\)

Key Concepts


Combinatorics

Check the Answer


Answer: $9$

AMC-10A (2013) Problem 7

Pre College Mathematics

Try with Hints


There are six programms: English, Algebra, Geometry, History, Art, and Latin. Since the student must choose a program of four course with the condition that there must contain English and at least one mathematics course. Therefore one course( i.e English) are already fixed and we have to find out the other subjects combinations…….

Can you now finish the problem ……….

There are Two cases :
Case 1: The student chooses both algebra and geometry.
This means that 3 courses have already been chosen. We have 3 more options for the last course, so there are 3 possibilities here.
case 2: The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by 2 , and then add to the previous.

Let us choose the mathematics course is algebra. so we can choose 2 of History, Art, and Latin, which is simply $3 \choose 2$=$3$. If it is geometry, we have another 3 options, so we have a total of 6 options if only one mathematics course is chosen.

can you finish the problem……..

Therefore the require ways are \(6+3=9\)

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Algebra Combinatorics Math Olympiad PRMO

Number of ways | PRMO 2017 | Question 9

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

Number of ways – PRMO 2017


There are five cities A, B, C, D, E on a certain island. Each city is connected to every other city by road, find numbers of ways can a person starting from city A come back to A after visiting some cities without visiting a city more than once and without taking the same road more than once. (The order in which he visits the cities such as A \(\rightarrow\) B \(\rightarrow\) C \(\rightarrow\) A and A \(\rightarrow\) C \(\rightarrow\) B \(\rightarrow\) A are different).

  • is 107
  • is 60
  • is 840
  • cannot be determined from the given information

Key Concepts


Number of ways

Integers

Combinatorics

Check the Answer


Answer: is 60.

PRMO, 2017, Question 9

Combinatorics by Brualdi

Try with Hints


First hint

A B C D E in this way orderwise such that from A person can visit B,C return to A in $4 \choose 2$ with 2! ways of approach

from A person visits B, C, D comes back to A in $4 \choose 3$ with 3! ways of approach

from A person visits B, C, D, E comes back to A in $4 \choose 4$ with 4! ways of approach

Second Hint

ways=\({4 \choose 2}(2!)+{4 \choose 3}(3!)+{4 \choose 4}(4!)\)

Final Step

=12+24+24

=12+48

=60.

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Combinatorics IIT JAM Statistics ISI M.Stat PSB ISI MSTAT

ISI MStat PSB 2010 Problem 2 | Combinatorics

This is a beautiful sample problem from ISI MStat PSB 2010 Problem 2. It’s all about how many isosceles triangle with sides of integers lengths one can construct under certain conditions. We provide a detailed solution with prerequisites mentioned explicitly.

Problem– ISI MStat PSB 2010 Problem 2

Find the total number of isosceles triangles such that the length of each side is a positive integer less than or equal to 40.

(Here equilateral triangles are also counted as isosceles triangle.)

Prerequisites

  • Basic counting principles.
  • Triangle inequality.

Solution

Let the sides of the isosceles triangle are k, k and l units, (length of the equal sides are k units)

So, our problem is all about finding triplets of form (k,k,l), where \(l, k \le 40\) . And how many such triplets can be found !

also since l,k and k are also sides of an (isosceles) triangle, so by triangle inequality,

\(l < 2k\) . Since l is an integer so, \(l \le 2k-1\)

So, we have \(l\le 40\) and \(l \le 2k-1\). combining we have \(l \le min(40, 2k-1)\) .

now let us define, \( l_k \) : possible number of l’s for a given k =1,2,….,40

here we must consider two cases.

Case-1 : when \(2k-1 \le 40 \Rightarrow k \le \lceil \frac{40+1}{2} \rceil = 20 \)

So, l can be any integer between 1 and 2k-1, since \( l\ \le 2k-1\) when \( k \le 20\)

so,there can be 2k-1 choices for the value of each k, when k=1,…,20.

Or, \( l_k = 2k-1 \) for k=1,2,….,20.

now, Case-2 : when 2k-1>40 \( \Rightarrow k > \lceil \frac{40+1}{2} \rceil = 20 \)

so, here l can take any integer between 1 to 40. So, there 40 choices for l for a given k, when k=21,22,…,40.

Or, \( l_k = 40 \) for k=21,22,….,40.

Combining Case-1 and Case-2, we have,

\(l_k = \begin{cases} 2k-1 & k=1,2,…,20 \\ 40 & k=21,22,…,40 \end{cases}\)

So, finally, let all possible number of triplets of form (k,k,l) are = \(T_40\)

So, \( T_{40} \) = \( \sum_{k=1}^{40}{l_k} \) = \( \sum_{k=1}^{20}{(2k-1)}\) + \( \sum_{k=21}^{40}{40} \) =\( (20 )^2 + 40 \times (40-20)=400+800= 1200. \)

So, \(T_{40}\) = 1200. hence we can find 1200 such isosceles triangles with sides of integer lengths such that the length of each sides are less than or equal to 40.

Food For Thought

Can you generalize this problem? i.e. can you find How many isosceles triangle one can construct, with sides of integer lengths, such that the sides are less than or equal to any integer N ? Can you find an elegant formula to express \(T_N \), for any integer N?

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Algebra Arithmetic Combinatorics I.S.I. and C.M.I. Entrance ISI Entrance Videos

Combinatorics and Integers | TOMATO B.Stat Objective 93

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Combinatorics and Integers.

Combinatorics and Integers (B.Stat Objective Question)


The highest power of 18 contained in \({50 \choose 25}\) is

  • 104
  • 1
  • 1154
  • none of these

Key Concepts


Integers

Combinatorics

Exponents

Check the Answer


Answer: 1

B.Stat Objective Problem 93

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

here \({50 \choose 25}\)=\(\frac{50!}{(25!)^{2}}\)=\(\frac{(50)(49)(….)(26)}{(25)(24)(…)(1)}\)

Second Hint

\(=(2)^{13}(49)(47)(45)(43)(41)(39)(37)(35)(33)(31)(29)(27) \times \frac{1}{12!}\)

Final Step

\(=(2)^{10}(49)(47)(15)(43)(41)(13)(37)(35)(11)(31)(29)\times \frac{1}{(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}[(2)^{3}(27)(9)(3)]\)

\(=(2)^{10}(49)(47)(15)(43)(41)(13)(37)(35)(11)(31)(29)\times \frac{1}{(12)(11)(10)(8)(7)(5)(4)(1)}[(2)(9)]\)gives a factor of \((18)^{1}\) then highest power of 18 is 1.

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Combinatorics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB Probability

ISI MStat 2019 PSA Problem 22 | Basic Probability

This is a problem from ISI MStat 2019 PSA Problem 22 based on basic probability principles.

Basic Probability – ISI MStat Year 2019 PSA Problem 22


A shopkeeper has 12 bulbs of which 3 are defective. She sells the bulbs by selecting them at random one at a time. 

What is the probability that the seventh bulb sold is the last defective one?

  • 3/44
  • 9/44
  • 13/44
  • 7/44

Key Concepts


combination

Basic Probability

Multiplication principle

Check the Answer


Answer: is 3/44

ISI MStat 2019 PSA Problem 22

A First Course in Probability by Sheldon Ross

Try with Hints


Let’s algorithmify the whole procedure.

Select positions of two defectives and the last defective being already fixed. This can be done in \( {6 \choose 2} \)

Let’s count the number of ways the defectives can sit once their places are fixed. This can be done in 3! ways .

Now the position of the non-defectives are fixed. 

Now let’s compute in how many ways they can sit there together in the remaining four places.

We have to arrange 9 non-defective objects in 4 places.

This can be done in 9 x 8 x 7 x 6 ways.

And without any restrictions we can select 7 bulbs in \( 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \) ways.

Hence the probability that the seventh bulb sold is the last defective one is \( \frac{ {6 \choose 2} \times 3! \times 9 x 8 x 7 x 6 }{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 } \)

Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Combinatorics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB Probability

ISI MStat 2018 PSA Problem 13 | Probability of functions

This is a beautiful problem from ISI MStat 2018 PSA problem 13 based on probability of functions. We provide sequential hints so that you can try .

Probability – ISI MStat Year 2018 PSA Problem 13


Consider the set of all functions from \( {1,2, \ldots, m} \) to \( {1,2, \ldots, n} \) where \( n>m .\) If a function is chosen from this set at random, what is the probability that it will be strictly increasing?

  • \( {n \choose m} / m^{n} \)
  • \( {n \choose m} / n^{m} \)
  • \( {{m+n-1} \choose m} / m^{n} \)
  • \( {{m+n-1} \choose m-1} / n^{m} \)

Key Concepts


combination

Check the Answer


Answer: is \( {n \choose m} / n^{m} \)

ISI MStat 2018 PSA Problem 13

A First Course in Probability by Sheldon Ross

Try with Hints


What is the total number of functions from \({1,2, \ldots, m}\) to \({1,2, \ldots, n}\) where (n>m)

You have to choose \(m\) numbers among \({1,2, \ldots, n}\) and assign it to the \({1,2, \ldots, m}\)
For each element of \({1,2, \ldots, m}\), there are \(n\) options from \({1,2, \ldots, n}\).
Hence \(n^m \) number of functions .

\(f(i) = a_i \)

The number of ways to select Select \(m\) elements among \({1,2, \ldots, n}\) is the same as the number of strictly ascending subsequences of length m taken from 1, 2, 3, …, n, which is the same as the number of subsets of size m taken from {1,2,3,…,n}, which is \( {n \choose m} \) .

Hence the probability that it will be strictly increasing \( \frac{ {n \choose m} }{ n^{m} } \)

Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Combinatorics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

ISI MStat 2018 PSA Problem 14 | All possible colorings

Try this problem from ISI MStat 2018 PSA Problem 14 based on all possible colorings. We provide sequential hints to help you solve the problem.

Probability – ISI MStat 2018 PSA 14


A flag is to be designed with 5 vertical stripes using some or all of the four colours: green, maroon, red and yellow. In how many ways can this be done so that no two adjacent stripes have the same colour?

  • 576
  • 120
  • 324
  • 432

Key Concepts


Basic counting principles

Check the Answer


Answer: is 324

ISI MStat 2018 PSA Problem 14

A First Course in Probability by Sheldon Ross

Try with Hints


How many ways you can colour the first stripe and the second stripe?

You have 4 colour options for the first stripe.
You have (4-1) = 3 colour options for the second stripe.
\( 4 \times 3 \).

How many ways you can colour the third stripe and the rest of the stripes?

You have (4-1) = 3 colour options for the second stripe.
You have (4-1) = 3 colour options for the fourth stripe.
You have (4-1) = 3 colour options for the fifth stripe.

Therefore, total number of ways to colour = \( (4 \times 3) \times 3 \times 3 \times 3 = 324 ) \)

Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Combinatorics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

ISI MStat 2019 PSA Problem 11 | Multiplication Principle

This is a beautiful problem from ISI MSTAT 2019 PSA problem 11 based on multiplication principles. We provide sequential hints so that you can try.

Multiplication Principle – ISI MStat 2019 PSA – 11


How many positive divisors of \( 2^55^3{11}^4 \) are perfect squares?

  • 60
  • 18
  • 120
  • 4

Key Concepts


Basic counting principles

Divisors of a number

Check the Answer


Answer: is 18

ISI MStat 2019 PSA Problem 11

A First Course in Probability by Sheldon Ross

Try with Hints


See in order to get the positive divisors of \( 2^55^3{11}^4 \) that are perfect squares , we need to take only the even powers of the primes {2,5,11}.

Now the maximum powers of 2 is 5 . So there are 3 choices for even powers {0,2,4} .

The maximum powers of 5 is 3 . So there are 2 choices for even powers {0,2}.

Now the maximum powers of 2 is 5 . So there are 3 choices for even powers {0,2,4} .

The maximum powers of 11 is 4 . So there are 3 choices for even powers {0,2,4}.

Hence by multiplication principle we have in total \( 3 \times 2 \times 3 =18 \) such positive divisors.

ISI MSTAT PSA and PSB Problems
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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AMC-8 Math Olympiad USA Math Olympiad

Probability in Divisibility | AMC-10A, 2003 | Problem 15

Try this beautiful problem from AMC 10A, 2003 based on Probability in Divisibility.

Probability in Divisibility – AMC-10A, 2003- Problem 15


What is the probability that an integer in the set \({1,2,3,…,100}\) is divisible by \(2\) and not divisible by \(3\)?

  • \(\frac {33}{100}\)
  • \(\frac{1}{6}\)
  • \(\frac{17}{50}\)
  • \(\frac{1}{2}\)
  • \(\frac{18}{25}\)

Key Concepts


Number system

Probability

divisibility

Check the Answer


Answer: \(\frac{17}{50}\)

AMC-10A (2003) Problem 15

Pre College Mathematics

Try with Hints


There are total number of integers are \(100\).and numer of integers divisible by \(2\) is \(\frac{100}{2}\)=\(50\). Now we have to find out divisible by \(2\) and not divisible by \(3\). so at first we have to find out the numbers of integers which are divisible by \(2\) and \(3\) both…….

can you finish the problem……..

To be divisible by both \(2\) and \(3\), a number must be divisible by the lcm of \((2,3)=6\).

Therefore numbers of integers which are divisible by \(6\)=\(\frac{100}{6}=16\) (between \(1\) & \( 100\))

can you finish the problem……..

Therefore the number of integers which are divisible by \(2\) and not divisible by \(3\)= \(50 – 16=34\).

So require probability= \(\frac{34}{100}=\frac{17}{50}\)

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Combinatorics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

ISI MStat 2019 PSA Problem 4 | Basic counting principle

This is a beautiful problem from ISI MSTAT 2019 PSA problem 4 based on basic counting principles. We provide sequential hints so that you can try.

Basic Counting Principles – ISI MStat 2019 PSA – 4


What is the number of 6 digit positive integers in which the sum of the digits is at least 52?

  • 66
  • 24
  • 28
  • 120

Key Concepts


Basic counting principles

Check the Answer


Answer: is

ISI MStat 2019 PSA Problem 4

A First Course in Probability by Sheldon Ross

Try with Hints


Find the Minimum Digit for each case of sum of the digits (S).

S = 54, Minimum Digit = 9

S = 53, Minimum Digit = 8

S = 54, Minimum Digit = 7 or 8

Let’s find the Second Minimum Digit and the Third Minimum Digit  for S = 53 and S = 52.

S = 53,
Second Minimum = 9

S = 52,
Minimum Digit = 7,
Second Minimum = 9

S = 52,
Minimum Digit = 8,
Second Minimum = 8,
Third Minimum = 9

Now it’s time for counting

S = 54
{999999}

S = 53,
{8,9,9,9,9,9} : Total = 6

S = 54,
{7,9,9,9,9,9} : Total = 6
{8,8,9,9,9,9} : Total = 15

Hence in total there are 1+6+6+15=28 such numbers .

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Outstanding Statistics Program with Applications

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