Indian Statistical Institute B.Math & B.Stat : Combinatorics

Problem 1: Points on a Curve

Question: A point \( P \) with coordinates \( (x,y) \) is said to be good if both \( x \) and \( y \) are positive integers. Find the number of good points on the curve \( xy=27027 \).

Solution:

First, note the prime factorization of the number: \( 27027 = 3^3 \times 13^1 \times 11^1 \times 7^1 \).

Since \( x \) and \( y \) must be positive integers, for every chosen positive integer \( x \) that divides 27027, there is exactly one corresponding integer \( y \) (where \( y = \frac{27027}{x} \)). Therefore, the number of good points is simply equal to the total number of positive divisors of 27027.

Using the divisor function formula \( \tau(n) \), if \( n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k} \), the number of divisors is \( (a_1 + 1)(a_2 + 1) \dots (a_k + 1) \).

Applying this to our factorization:

\[ \tau(27027) = (3+1)(1+1)(1+1)(1+1) = 4 \times 2 \times 2 \times 2 = 32 \]

There are 32 good points on the curve.

Problem 2: Ordered Triplets

Question: What is the number of ordered triplets \( (a, b, c) \), where \( a, b, c \) are positive integers (not necessarily distinct), such that \( abc = 1000 \)?

Solution:

First, find the prime factorization of 1000, which is \( 2^3 \times 5^3 \).

Any ordered triplet \( (a, b, c) \) satisfying the condition must be of the form:

• \( a = 2^{l_1} \times 5^{m_1} \)
• \( b = 2^{l_2} \times 5^{m_2} \)
• \( c = 2^{l_3} \times 5^{m_3} \)

For \( abc = 2^3 \times 5^3 \) to hold true, the powers of each prime base must add up to 3. This gives us two independent equations:

1. \( l_1 + l_2 + l_3 = 3 \)
2. \( m_1 + m_2 + m_3 = 3 \)

We need to find the number of non-negative integer solutions for each equation. Using the "Stars and Bars" combinatorics method, the number of solutions to \( x_1 + x_2 + x_3 = n \) is given by \( \binom{n+r-1}{r-1} \), where \( r \) is the number of variables.

For the powers of 2: \( \binom{3+3-1}{3-1} = \binom{5}{2} = 10 \) solutions.

For the powers of 5: \( \binom{3+3-1}{3-1} = \binom{5}{2} = 10 \) solutions.

Since the distributions of the powers of 2 and 5 are independent, we multiply the possibilities:

\[ 10 \times 10 = 100 \]

Thus, in total 100 ordered triplets are possible.

Problem 3: Derangements and Boxes

Question: There are 8 balls numbered \( 1,2,\dots,8 \) and 8 boxes numbered \( 1,2,\dots,8 \). Find the number of ways one can put these balls in the boxes so that each box gets one ball and exactly 4 balls go into their corresponding numbered boxes.

Solution:

First, we must choose which 4 balls will go into their correct boxes. This can be done in \( \binom{8}{4} \) ways. For any such selection—say we choose \( \{Ball_2, Ball_5, Ball_3, Ball_7\} \)—there is exactly 1 way for them to go into their corresponding boxes.

Now, the remaining 4 balls—in this example, \( \{Ball_1, Ball_4, Ball_6, Ball_8\} \)—must be placed into the remaining 4 boxes such that none of them end up in a box matching their own number. This is the classic definition of a derangement.

The number of ways to derange \( n \) objects is denoted as \( D_n \). The formula for a derangement is:

\[ D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + (-1)^n\frac{1}{n!} \right) \]

For our remaining 4 balls, we calculate \( D_4 \):

\[ D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 12 - 4 + 1 = 9 \]

The total number of valid configurations is the product of our choices:

\[ \text{Total Ways} = \binom{8}{4} \times D_4 = 70 \times 9 = 630 \]

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