ISI Entrance 2011 - B.Math Subjective Paper| Problems & Solutions

Here, you will find all the questions of ISI Entrance Paper 2011 from Indian Statistical Institute's B. Math Entrance. You will also get the solutions soon of all the previous year problems.

Problem 1 :

Let $a \geq 0$ be a constant such that $\sin (\sqrt{x+a})=\sin (\sqrt{x})$ for all $x \geq 0 .$ What can you say about $a$ ? Justify your answer.

Problem 2 :

Let $f(x)=e^{-x}$ for $x>0$. Define a function $g$ on the nonnegative real numbers as follows: for each integer $k>0$, the graph of the function $g$ on the interval $[k, k+1]$ is the straight line segment connecting the points $(k, f(k))$ and $(k+1, f (k+1) )$. Find the total area of the region which lies between the curves of $f$ and $g$.

Problem 3 :

For any positive integer $n$, show that $\frac{1}{2}\cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \cdot \frac{(2 n-1)}{2 n}<\frac{1}{\sqrt{2 n+1}}$.

Problem 4 :

If $a_{1}, \ldots, a_{7}$ are not necessarily distinct real numbers such that $1<a_{i}<13$ for all $i$, then show that we can choose three of them such that they are the lengths of the sides of a triangle.

Problem 5 :

For any real number $x,$ let $[x]$ denote the largest integer which is less than or equal to $x$. Let $N_{1}=2, N_{2}=3, N_{3}=5, \ldots$ be the sequence of non-square positive integers. If the $n$ th non-square positive integer satisfies $m^{2}<N_{n}<$ $(m+1)^{2},$ then show that $m=\left[\sqrt{n}+\frac{1}{2}\right]$

Problem 6 :

Let $R$ and $S$ be two cubes with sides of lengths $r$ and $s$ respectively, where $r$ and $s$ are positive integers. Show that the difference of their volumes equals the difference of their surface areas, if and only if $r=s$.

Problem 7 :

Let $A B C$ be any triangle and let $O$ be a point on the line segment $B C .$ Show that there exists a line parallel to $A O$ which divides the triangle $A B C$ into two equal parts of equal area.

Problem 8 :

Let $t_{1}<t_{2}<\cdots<t_{99}$ be real numbers, and consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x)=\left|x-t_{1}\right|+\left|x-t_{2}\right|+\cdots+\left|x-t_{99}\right| .$ Show that $\min _{x \in R} f(x)=f\left(t_{50}\right)$

Some useful link :

ISI Entrance 2009 - B.Math Subjective Paper| Problems & Solutions

Here, you will find all the questions of ISI Entrance Paper 2009 from Indian Statistical Institute's B. Math Entrance. You will also get the solutions soon of all the previous year problems.

Problem 1:

Let $x, y, z$ be non-zero real numbers. Suppose $\alpha, \beta, \gamma$ are complex numbers such that $|\alpha|=|\beta|=|\gamma|=1 .$ If $x+y+z=0=\alpha x+\beta y+\gamma z,$ then prove that $\alpha=\beta=\gamma$

Problem 2:

Let $c$ be a fixed real number. Show that a root of the equation
$$
x(x+1)(x+2) \cdots(x+2009)=c
$$
can have multiplicity at most 2. Determine the number of values of $c$ for which the equation has a root of multiplicity 2 .

Problem 3:

Let $1,2,3,4,5,6,7,8,9,11,12, \ldots$ be the sequence of all the positive integers integers which do not contain the digit zero. Write $\{a_{n}\}$ for this sequence. By comparing with a geometric series, show that $\sum_{n} \frac{1}{a_{n}}<90 .$

Problem 4:

$$
\begin{aligned}
&\text { Find the values of } x, y \text { for which } x^{2}+y^{2} \text { takes the minimum value where }(x+\
&5)^{2}+(y-12)^{2}=14
\end{aligned}
$$

Problem 5:

Let $p$ be a prime number bigger than $5 .$ Suppose, the decimal expansion of $1 / p$ looks like $0 . \overline{a_{1} a_{2} \cdots a_{r}}$ where the line denotes a recurring decimal. Prove that $10^{r}$ leaves a remainder of 1 on dividing by $p$.

Problem 6:

Let $a, b, c, d$ be integers such that $a d-b c$ is non-zero. Suppose $b_{1}, b_{2}$ are integers both of which are multiples of $a d-b c .$ Prove that there exist integers simultaneously satisfying both the equalities $a x+b y=b_{1}, c x+d y=b_{2}$.

Problem 7:

Compute the maximum area of a rectangle which can be inscribed in a triangle of area $M$

Problem 8:

Suppose you are given six colours and, are asked to colour each face of a cube by a different colour. Determine the different number of colourings possible.

Problem 9:

Let $f(x)=a x^{2}+b x+c$ where $a, b, c$ are real numbers. Suppose $f(-1), f(0), f(1) \in$ $[-1,1] .$ Prove that $|f(x)| \leq 3 / 2$ for all $x \in[-1,1]$

Problem 10:

Given odd integers $a, b, c,$ prove that the equation $a x^{2}+b x+c=0$ cannot have a solution $x$ which is a rational number.

Some useful link :

Likelihood & the Moment | ISI MStat 2016 PSB Problem 7

This problem is a beautiful example when the maximum likelihood estimator is same as the method of moment estimator. Infact, we have proposed a general problem, is when exactly, they are equal? This is from ISI MStat 2016 PSB Problem 7, Stay Tuned.

Problem

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed random variables ~ \(X\) with probability mass function
$$
f(x ; \theta)=\frac{x \theta^{x}}{h(\theta)} \quad \text { for } x=1,2,3, \dots
$$
where \(0<\theta<1\) is an unknown parameter and \(h(\theta)\) is a function of \(\theta\) Show that the maximum likelihood estimator of \(\theta\) is also a method of moments estimator.

Prerequisites

Solution

This \(h(\theta)\) looks really irritating.

Find the \( h(\theta) \).

\( \sum_{x = 1}^{\infty} f(x ; \theta) = \sum_{x = 1}^{\infty} \frac{x \theta^{x}}{h(\theta)} = 1 \)

\( \Rightarrow h(\theta) = \sum_{x = 1}^{\infty} {x \theta^{x}} \)

\( \Rightarrow (1 - \theta) \times h(\theta) = \sum_{x = 1}^{\infty} {\theta^{x}} = \frac{\theta}{1 - \theta} \Rightarrow h(\theta) = \frac{\theta}{(1 - \theta)^2}\).

Maximum Likelihood Estimator of \(\theta\)

\( L(\theta)=\prod_{i=1}^{n} f\left(x_{i} | \theta\right) \)

\( l(\theta) = log(L(\theta)) = \sum_{i=1}^{n} \log \left(f\left(x_{i} | \theta\right)\right) \)

Note: All irrelevant stuff except the thing associated with \( \theta \) is kept as constant (\(c\)).

\( \Rightarrow l(\theta) = c + n\bar{X}log(\theta) - nlog(h(\theta)) \)

\( l^{\prime}(\theta) = 0 \overset{Check!}{\Rightarrow} \hat{\theta}_{mle} = \frac{\bar{X} -1}{\bar{X} +1}\)

Method of Moments Estimator

We need to know the \( E(X)\).

\( E(X) = \sum_{x = 1}^{\infty} xf(x ; \theta) = \sum_{x = 1}^{\infty} \frac{x^2 \theta^{x}}{h(\theta)} \).

\( E(X)(1 - \theta) = \sum_{x = 1}^{\infty} \frac{(2x-1)\theta^{x}}{h(\theta)} \).

\( E(X)\theta(1 - \theta) = \sum_{x = 1}^{\infty} \frac{(2x-1)\theta^{x+1}}{h(\theta)} \)

\( E(X)((1 - \theta) - \theta(1 - \theta)) =\frac{\sum_{x = 1}^{\infty} 2\theta^{x} - \theta }{h(\theta)} = \frac{\theta(1 + \theta)}{(1 - \theta)h(\theta)}\).

\( \Rightarrow E(X) = \frac{\theta(1 + \theta)}{(1 - \theta)^3h(\theta)} = \frac{1+\theta}{1-\theta}.\)

\( E(X) = \bar{X} \Rightarrow \frac{1+\theta_{mom}}{1-\theta_{mom}}= \bar{X} \Rightarrow \hat{\theta}_{mom} = \frac{\bar{X} -1}{\bar{X} +1}\)

Food For Thought and Research Problem

Normal (unknown mean and variance), exponential, and Poisson all have sufficient statistics equal to their moments and have MLEs and MoM estimators the same (not strictly true for things like Poisson where there are multiple MoM estimators).

So, when do you think, the Method of Moments Estimator = Maximum Likelihood Estimator?

Pitman Kooper Lemma tells us that it is an exponential family.

Also, you can prove that that there exists a specific form of the exponential family.

Stay tuned for more exciting such stuff!

Inequality Module for I.S.I. Entrance and Math Olympiad

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Start Date: 13th October, 2019

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Class timing: Sunday, 8:30 PM to 10 PM.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Faculty: Srijit Mukherjee, B.Stat (2019), M.Stat (2021), Indian Statistical Institute" open="on" _builder_version="4.0"]Research Interests: Probability Theory, Dynamical Systems, Discrete Mathematics

[/et_pb_accordion_item][et_pb_accordion_item title="Course pre-requisites" _builder_version="4.0" open="off"]Algebraic Manipulation
Trigonometry, Euclidean Geometry[/et_pb_accordion_item][et_pb_accordion_item title="Target Exams" _builder_version="4.0" open="off"]Pre-RMO, RMO, INMO, IMO ( India Olympiad )

ISI – CMI Entrance ( India Undergraduate Entrance )

AMC 8, AMC 10,  AMC 12, AIME, SMO, etc ( USA Olympiad, Non – Indian Olympiad )[/et_pb_accordion_item][et_pb_accordion_item title="Watch the video" _builder_version="4.0" open="off"]

https://youtu.be/CdvZFbm2G0g[/et_pb_accordion_item][/et_pb_accordion][et_pb_button button_url="https://cheenta.com/contact-us" url_new_window="on" button_text="Join" button_alignment="center" _builder_version="4.0" custom_button="on" button_text_size="48px" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" custom_margin="50px||||false|false" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][/et_pb_column][/et_pb_row][/et_pb_section]

AM GM inequality in ISI Entrance

Arithmetic Mean and Geometric Mean inequality form a foundational principle. This problem from I.S.I. Entrance is an application of that.

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ISI Entrance Solutions
I.S.I. Entrance Program

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How are Bezout's Theorem and Inverse related? - Number Theory

The inverse of a number (modulo some specific integer) is inherently related to GCD (Greatest Common Divisor). Euclidean Algorithm and Bezout's Theorem forms the bridge between these ideas. We explore them in a very lucid manner.

How to use Invariance in Combinatorics - ISI Entrance Problem

Invariance is a fundamental phenomenon in mathematics. In this combinatorics problem from ISI Entrance, we discuss how to use invariance.

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Mathematical Circles Inequality Problem

[et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]If \(a,b\) are positive reals such that \(a+b<2\) ,then prove that  $$ \displaystyle \frac {1}{1+a^2} + \frac {1}{1+b^2} \leq \frac {2}{1+ab} $$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Mathematical Circles[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Inequality involving AM-GM[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Mathematical Circles[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Assume that the given inequality is true for \( a>0 , b>0 \) and \( a+b<2 \) . Then proceed .[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]In order to simplyfy the given inequality multiply both the sides by \( (1+a^2)(1+b^2)(1+ab) \) (as its a positive quantity and it is directly coming from   \( a>0 , b>0\)  ) .    [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]Come up with the simplest form of inequality  i.e. \( (a-b)^2 (1-ab) \geq 0 \) .[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]\( (a-b)^2 \geq 0 \) as \( a,b \in{R}\) . And to get \( (1-ab)>0 \) use the well known inequality for positive reals i.e. \( AM \geq GM \) and the still unused inequality i.e \( a+b <2 \) also . $$ \displaystyle a>0 , b>0 \Rightarrow \sqrt{ab}>0 \Rightarrow( 1+ \sqrt{ab})>0  \\  a>0 , b>0 , a+b <2 \Rightarrow  1 > \frac{a+b}{2} \geq \sqrt {ab} \\ \Rightarrow 1 > \sqrt{ab} \\ \Rightarrow ( 1 - \sqrt{ab}) >0 \\ \Rightarrow (1 - \sqrt{ab}) (1+ \sqrt{ab}) >0 \\ \Rightarrow (1 - ab)>0 $$[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="7" _address="0.1.0.7"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="Math Olympiad Program" url="https://cheenta.com/matholympiad/" url_new_window="on" image="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://cheenta.com/matholympiad/" link_option_url_new_window="on" _i="8" _address="0.1.0.8"]

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="9" _address="0.1.0.9"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="10" _address="0.1.0.10"]

Similar Problems

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Application of eigenvalue in degree 3 polynomial: ISI MMA 2018 Question 14

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]Let A be a 3 x 3 real matrix with all the diagoanl entries equal to 0 . If 1 + i is an eigen value of A , the determinant of A equal ?[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Sample Questions (MMA ) :2019[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Schaum's Outline of Linear Algebra[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Let A be a 3 x 3 real matrix with trace 0 Now( 1+i) be an eigen value . (Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values, proper values, or latent roots & Eigen vectors are In linear algebra,) (An eigenvector  or characteristic vector of a linear transformation is a non-zero vector that changes by only a scalar factor when that linear transformation is applied to it. More formally, if T is a linear transformation from a vector space V over a field F into itself and v is a vector in V that is not the zero vector, then v is an eigenvector of T if T(v) is a scalar multiple of v. This condition can be written as the equation ) [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]So ,( 1 + i) be the roots of the characteristic poly of A  Now A is a real matrix, char poly of A  \(\epsilon \mathbb{R}[x]\) [Right!] Therefore ( 1 - i) is also a root of char poly of A[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]deg( char poly of A ) =3 So , it has two imaginary roots & one real root Let real root be r Note tr(A) = 0 => r + 1 +i +1 -i = 0 , => r = -2 Can you play with the determinants now ?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]We know the determinant of A is the product of all eigen values  (-2)(1+i)(1-i) = detA => det(A) = -2[1 +1] = -4[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

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Coloring problems: ISI MMA 2018 Question 10

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[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]A new flag of ISI club is to be designed with 5 vertical strips using some or all the four colours : green , naroon , red and yellow . In how many ways this can be done  so that no two adjacent strips have the same colour ? [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Sample Questions ( MMA ) :2019 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Combinatorics [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Schaum's outline of combinatorics by  Balakrishnan [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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This is an application of multiplication property in combinatorics. The rule of product or multiplication principle is a basic counting principle (a.k.a. the fundamental principle of counting). Stated simply, it is the idea that if there are a ways of doing something and b ways of doing another thing, then there are a · b ways of performing both actions.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]ISI club has 5 vertical strips . We have to colour them using 4 colours . So , the first strip can be coloured in 4 ways . WLOG we take it to be green . Can the second strip be coloured green ?  No  ! Right ? [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]So , we have to choose the second strip from rest of the colours . [ Because two adjacent strip has same colour ] [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]Similarly , third strip can be coloured into 3 ways , fourth strips can be coloured into 3 ways and fifth strips can be coloured into 3 ways .[ We have to exclude the colour of the second one]  Therefore , the total number  of probabilities are - \(3^4\) x 4 = 324 [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

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