# Algebra and Combination | AIME I, 2000 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebra and Combination.

## Algebra and combination - AIME 2000

In expansion $$(ax+b)^{2000}$$ where a and b are relatively prime positive integers the coefficient of $$x^{2}$$ and $$x^{3}$$ are equal, find a+b

• is 107
• is 667
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Combination

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

here coefficient of $$x^{2}$$= coefficient of $$x^{3}$$ in the same expression

then $${2000 \choose 1998}a^{2}b^{1998}$$=$${2000 \choose 1997}a^{3}b^{1997}$$

then $$b=\frac{1998}{3}$$a=666a where a and b are relatively prime that is a=1,b=666 then a+b=666+1=667.

.

# 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

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 $$\frac{ {6 \choose 2} \times 3! \times 9 x 8 x 7 x 6 }{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 }$$

# 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

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 $$\frac{ {n \choose m} }{ n^{m} }$$

# ISI MStat PSA 2019 Problem 6 | Basic Counting principles

This is a beautiful problem from ISI MSTAT 2019 PSA problem 6 based on basic counting principles. We provide sequential hints so that you can try.

## Basic Counting Principles - ISI MStat Year 2019 PSA - 6

How many times does the digit ‘2’ appear in the set of integers $$\{1,2,…,1000\}$$?

• 590
• 600
• 300
• 299

### Key Concepts

Basic counting principles

ISI MStat 2019 PSA Problem 6

A First Course in Probability by Sheldon Ross

## Try with Hints

Let’s count the number of times 2 occurs once.

_ _ _

The position of 2 can be selected in 3 ways. The rest in 9 x 9.

Total number of 2 = 1 x 3 x 9 x 9 = 243

Let’s count the number of times 2 occurs twice.

_ _ _

The positions of 2 can be selected in 3 ways. The rest in 9.

Total number of 2 = 2 x 3 x 9  = 54

Let’s count the number of times 2 occurs thrice.

The positions of 2 can be selected in 1 way. The rest in 1.

Total number of 2 = 3 x 1 x 1  = 3

Total Number of Such Numbers = 243 + 54 + 3 = 300

# Let's Permute | ISI MStat 2018 PSB Problem 3

This problem is an easy application of the basic algorithmic ideas to approach a combinatorics problem using permutation and combination and basic counting principles. Enjoy this problem 3 from ISI MStat 2018 PSB.

## Problem

Consider all permutations of the integers $$1,2, \ldots, 100$$ . In how many
of these permutations will the $$25^{th}$$ number be the minimum of the
first 25 numbers and the $$50^{th}$$ number be the minimum of the first 50
numbers?

## Solution

Whenever you are counting, think in terms of algorithms or steps in which you will do the task given in the problem and count each of the steps. Now the conjunction in between the steps ( and $$\cap$$ , or $$\cup$$ ) will show you the way.

#### Steps

Select the first 50 numbers. Fix the minimum element. Out of the 49 other elements, select the first 25 numbers. Fix the minimum element of these 25 numbers. Then permute the rest.

AND for each of these selections of two sets (Multiplication Principle),

#### Step 2

Once you select the three sets, the minimum of each of these sets is already fixed. So, for each of these sets, you have to permute the other first 24 numbers (1 - 24) and the next 24 numbers (from 26 - 49) and the other 50 numbers (50 - 100).

You can do that in $$24! \times 24! \times 50!$$ ways.

Therefore, you can do the whole task in $${100 \choose 50} \times { 49 \choose 25 } \times 24! \times 24! \times 50!$$ ways.

## Food for Thought

• Do the problem for $$2n$$ numbers.
• Do the problem if you want to fix the minimum $$k$$ numbers in each set out of $$2n$$ numbers.
• Do the problem if you want to fix the minimum $$n$$ numbers in each set out of $$2n$$ numbers. Does it match your expectation?
• Generalize and Degeneralize and make your own problem out of this. Comment it below. We will update it here if we think the problem is appropriate.

# Non-Consecutive Selection | ISI MStat 2019 PSB Problem 3

This problem is a beautiful and simple application of the bijection principle to count the number of non-consecutive selection of integers in combinatorics from Problem 3 of ISI MStat 2019 PSB.

## Problem - Non-Consecutive Selection

Elections are to be scheduled for any seven days in April and May. In how many ways can the seven days be chosen such that elections are not scheduled on two consecutive days?

### Prerequisites

• $$a < b$$, then $$a$$ and $$b+1$$ are never consecutive.
• Combination ( Choose Principle )
• Bijection Principle

## Solution

The problem is based on the first prerequisite mainly. That idea mathematicalizes the problem.

Out of the 61 days in April and May, we have to select 7 non-consecutive days. Let's convert this scenario to numbers.

Out of {$$1, 2, 3, ... , 61$$}, we have to select 7 non-consecutive numbers.

#### Lemma

$$y_1 < y_2 < y_3 < ... < y_7$$ are 7 non-consecutive numbers $$\iff$$ $$y_i$$ is of the form $$x_i + (i-1)$$ where $$x_1 < x_2 < x_3 < ... < x_7$$.

For example

You select {1, 3, 4, 5, 6, 8}. You change it to {1, 3 + 1, 4 + 2 , 5 + 3, 6 + 4 , 8 + 5} = { 1, 4, 6, 8, 10 , 13}, which are never consecutive.

Essentially, we are counting the non-consecutive integers in a different way, which helps us to count them.

So, we have to choose $$x_1 < x_2 < x_3 < ... < x_7$$, where the maximum $$x_7 + (7-1) = x_7 + 6 \leq 61 \Rightarrow x_7 \leq 55$$.

Hence, the problem boiled down to choosing $$x_1 < x_2 < x_3 < ... < x_7$$ from {$$1, 2, 3, ... , 55$$}, which is a combination problem.

We can have to just choose 7 such numbers. The number of ways to do so is $${55}\choose{7}$$.

# Time and Work | PRMO-2017 | Problem 3

Try this beautiful problem from PRMO, 2017 based on Time and work.

## Time and work | PRMO | Problem-3

A contractor has two teams of workers : team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins team A after four days. The team A withdraws after two more days. For how many more days should team B work to complete the job ?

• $20$
• $16$
• $13$

### Key Concepts

Arithmetic

multiplication

unitary method

Answer:$16$

PRMO-2017, Problem 3

Pre College Mathematics

## Try with Hints

In the problem,we notice that first 4 days only A did the work.so we have to find out A's first 4 days work done.next 2 days (A+B) did the work together,so we have to find out (A+B)'s 2 days work.

so we may take the total work =1

A's 1 day's work= $$\frac{1}{12}$$ and B's 1 day's work=$$\frac{1}{36}$$

Can you now finish the problem ..........

Now B did complete the remaining work.so you have to find out the remaining work and find out how many more days taken....

so to find the remaining work subtract (A's 4 day;s work + (A+B)'S 2 days work)) from the total work

Can you finish the problem........

Let the total work be 1

A can complete the total work in 12 days,so A'S 1 day's work=$$\frac{1}{12}$$

B can complete the total work in 36 days, so B's 1 day's work=$$\frac{1}{36}$$

First 4 days A's workdone=$$\frac{4}{12}=\frac{1}{3}$$

After 4 days B joined and do the work with A 2 days

So $$(A+B)$$'s 2 day's workdone=$$2 \times( \frac{1}{12}+\frac{1}{36})$$=$$\frac{2}{9}$$

Remaining workdone=$$(1-\frac{1}{3}-\frac{2}{9}$$)=$$\frac{4}{9}$$

B will take the time to complete the Remaining work=$$36 \times \frac{4}{9}$$=16

Hence more time taken=16

# Combination of Cups | PRMO-2018 | Problem 11

Try this beautiful problem from PRMO, 2018 based on Combination of Cups.

## Combination of Cups | PRMO | Problem 11

There are several tea cups in the kitchen, some with handle and the others without handles. The
number of ways of selecting two cups without a handle and three with a handle is exactly 1200.
What is the maximum possible number of cups in the kitchen ?

• $24$
• $29$
• $34$

### Key Concepts

Algebra

combination

maximum

Answer:$29$

PRMO-2018, Problem 11

Pre College Mathematics

## Try with Hints

We assume that the number of cups with handle=m and number of cups without handle =n

Can you now finish the problem ..........

The way of select 3 cups with a handle from m cups = $$m_{C_3}$$ and select 2 cups without a handle from n cups =$$n_{C_2}$$

Can you finish the problem........

We assume that the number of cups with handle=m and number of cups without handle =n

The way of select 3 cups with a handle from m cups = $$m_{C_3}$$ and select 2 cups without a handle from n cups =$$n_{C_2}$$

Therefore $$m_{C_3} \times n_{C_2}$$=1200

$$\Rightarrow \frac {m(m-1)(m-2)}{6} \times \frac{n(n-1)}{2}$$=$$2^4 \times 3 \times 5^2$$

$$m=4,y=25$$ and $$m=5,n=16$$

Therefore $$m+n$$ be maximum i.e $$m=4,n=25$$

Maximum possibile cups=25+4=29

# 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

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=$$\frac{2}{5}$$

# 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?

• $$\frac{2}{3}$$
• $$\frac{1}{3}$$
• $$\frac{1}{4}$$

### Key Concepts

probability

Equilly likely

Number counting

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.......

$$1 - \frac{2}{4} \times \frac{2}{3}$$= $$1 - \frac{1}{3} = \frac{2}{3}$$