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## 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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## Problem – Numbers and Group (B.Stat Objective Problem)

We are going to discuss about Numbers and Group from B.Stat Objective Problem .

In a group of 120 persons there are 80 Bengalis, 40 Gujratis, Further 71 persons in the group are Muslims and the remaining Hindus, Then the number of Bengali Muslims in the group is

• 8
• more than 30
• 11
• 10

### Key Concepts

number Series

Algebra

Integers

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

Assuming that out of 80 Bengalis, when 49 are Hindus then 31 are Muslims and when 9 are Hindus 71 are Muslims

Second Hint

Out of 71 Muslims when 40 are Gujratis 31 are Bengalis else 71 are Bengalis which contradicts with other options

Final Step

Then number of Bengali Muslims are from 31 to 71 that is more than 30.

Categories

## Logic and Group | TOMATO B.Stat Objective Question

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Logic and Group.

## Logic and Group ( B.Stat Objective Question )

Four passengers in a compartment of the Delhi-Howrah Rajdhani Express discover that they form an interesting group. Two are lawyers and two are doctors. Two of them speak Bengali and the other two Hindi and no two of the same profession speak the same language . They also discover that two of them are Christians and two Muslims, no two of the same religion are of the same profession and no two of the same religion speak the same language. The Hindi speaking doctor is a christian.Then which one holds

• BLM
• CLB
• BDC
• BDH

### Key Concepts

number Series

Logic

Integers

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

one doctor HDC then other doctors by given conditions is BDM

Second Hint

then other two LM and LC and for DMB then LM is H that is HLM

Final Step

Then LC is B that is BLC or CLB.

Categories

## Box and ball Probability | B.Stat Objective TOMATO Problem 59

Try this problem from I.S.I. B.Stat TOMATO Objective Problem based on Box and ball Probability.

## Box and ball Probability ( B.Stat Objective Problem )

A box contains 100 balls of different colours 28 red 17 blue 21 green 10 white 12 yellow 12 black. The smallest number n such that any n balls drawn from the box will contain at least 15 balls of the same colour is

• 8
• 77
• 11
• 10

### Key Concepts

number Series

Logic

Integers

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

28 -15 gives 13 and 17-15 gives 2 and 21 -15 gives 6 rest are less than 15

Second Hint

then 100 -13=77 and 100-2=98 and 100-6= 94

Final Step

then smallest number= 77