Refer the blog questions a few days before:

**Question 1:**

Let be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

**Answer 1:**

Without loss of generality, we may take …..call this relation (i).

Let, if possible, no three of the given numbers be the lengths of the sides of a triangle. (That is, no three satisfy the triangle inequality. Note that when we say three numbers a, b and c satisfy the triangle inequality —- it means all the following three inequalities have to hold simultaneously: , and ). We will consider triplets and . As these numbers do not form the lengths of the sides of a triangle, the sum of the smallest two numbers should not exceed the largest number, that is, . Hence, we get the following set of inequalities:

gives giving .

gives giving

gives giving

gives giving

gives giving

gives giving

gives giving

gives giving

contradicting the basic hypothesis. Hence, there exists three numbers among the given numbers which form the lengths of the sides of a triangle.

**Question 2:**

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide the collection into two parts such that each person has at most 1 enemy in his sub-collection.

**Answer 2:**

Let C denote the collection of given 1234 persons. Let be a partition of C. Let denote the total number of enemy pairs in . Let denote the total number of enemy pairs in .

Let denote the total number of enemy pairs corresponding to the partition of C. Note is an integer greater than or equal to zero. Hence, by Well-Ordering Principle, there exists a partition having the least value of .

Claim: This is “the” required partition.

Proof: If not, without loss of generality, suppose there is a person P in having at least 2 enemies in . Construct a new partition of C as follows: and . Now, . Hence, contradicting the minimality of . QED.

**Problem 3:**

A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw.

What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any is repeated?

**Answer 3:**

The total number of possible outcomes is . To find the total number of favourable outcomes we proceed as follows:

Let a be any odd integer such that and let us count the sequences having a as least element.

(i) There is only one sequence with a repeated thrice.

(ii) There are sequences of the form with . For each such sequence there are three distinct permutations possible. Hence, there are in all sequences with a repeated twice.

iii) When , for values of a satisfying , sequences of the form with are possible and the number of such sequences is . For each such sequence, there are six distinct permutations possible. Hence, there are sequences in this case.

Hence, for odd values of a between 1 and , the total counts of possibilities , , in the above cases are respectively.

.

Hence, the total number A of favourable outcomes is . Hence, the required probability is . QED>

Cheers,

Nalin Pithwa