Problem 16 - Algebra for JEE ADVANCED

Question

A batsman scores exactly a century by hitting fours and sixes in twenty consecutive balls. In how many different ways can he do it if some balls may not yield runs and the order of boundaries and overboundaries are taken into account?

Step-by-Step Solution

Answer

Calculate feasible (x, y) pairs, then sum their permutations.

Step 1: Understanding the Problem

We are given a scenario where a batsman scores exactly 100 runs using only fours (4 runs) and sixes (6 runs) over twenty balls. No other runs and no extras like singles, doubles, or threes are involved. We need to find out how many different sequences of fours, sixes, and potentially 'dot' balls (zero runs) can give exactly 100 runs in 20 deliveries, with the order of runs being significant.

Step 2: Introducing Variables

Let's denote the number of balls on which the batsman scores fours as \( x \) and the number of balls on which he scores sixes as \( y \). Another variable, \( z \), will represent the balls that yield no runs (dot balls). Hence, the main contributing equation is: \( 4x + 6y + 0z = 100 \). Also, the total number of balls is limited to 20, so \( x + y + z = 20 \).

Step 3: Eliminating Unnecessary Variable

Using the equation \( x + y + z = 20 \), we can express \( z \) in terms of \( x \) and \( y \): \( z = 20 - (x + y) \). Substitute \( z \) in the main equation: \( 4x + 6y = 100 \). These two equations are the constraints that will be used to solve for the possible combinations of \( x \) and \( y \).

Step 4: Finding Feasible Solutions

To solve \( 4x + 6y = 100 \), iterate possible integer values for \( y \) (from 0 to 16) and solve for \( x \) using \( x = \frac{100 - 6y}{4} \). Simultaneously check \( x + y \leq 20 \) to ensure the number of balls doesn't exceed 20.

Step 5: Alternative Calculation Approach

Given simplicity by factoring the main equation further or checking for bounds initially might skip some computational steps. Checking inequalities: \( 4x + 6y \leq 100 \) often retrieves possible boundary values. Analyze valid ranges for \( y \) till \( x \) becomes non-negative and ensure whole numbers do not exceed the total number of balls.

Step 6: Using Permutations for Order

Once feasible (\( x, y, z \)) solutions are found, calculate the permutations for each set using \( \frac{20!}{x!y!z!} \). This formula gives the number of ways to arrange these solutions, acknowledging each type of run (or lack thereof) as indistinct among their peers, but distinct across type.

Step 7: Summing All Permutations

For each valid combination of (\( x, y, z \)) that satisfies both the score and ball conditions, calculate the respective permutations and sum them all to get the total number of different ways runs can be achieved over 20 balls.

Key Concepts

Permutations

Permutations are an essential concept in combinatorics, dealing with the arrangement of elements in a specific order. In this exercise, we need to calculate the permutations of scores, including fours, sixes, and dot balls, over 20 deliveries. This is crucial because each arrangement leads to a different valid sequence in the context of cricket scoring.

A permutation considers not just which elements are used, but their exact order. To compute permutations when certain elements repeat, we use the formula:

• \( \frac{n!}{x!y!z!} \)

Here, \(n\) is the total number of items to arrange (20 balls in this exercise), while each denominator signifies the factorial of occurrences of each specific element: fours (\(x\)), sixes (\(y\)), and dot balls (\(z\)). This formula accounts for indistinguishable arrangements within identical sets (e.g., two fours have no unique order between themselves).

The combinatorial structure of this problem emphasizes understanding how permutations function when factorized out to real-world scenarios like cricket.

Integer Solutions

Finding integer solutions to equations is about specifying the whole number values that satisfy a given equation. In our exercise, we are looking at the equation:

• \( 4x + 6y = 100 \)

This equation arises because the batsman scores only with fours and sixes, and they total 100 runs. The challenge is to find combinations of integers \(x\) and \(y\) that satisfy this equation alongside another constraint involving the number of balls:

• \( x + y \leq 20 \)

To solve this type of problem, a systematic approach is needed.

We can test feasible values of \(y\), ranging from 0 to a maximum where conditions hold true. For each \(y\), we solve for \(x\) such that \(x\) remains an integer and fits the ball constraint. Checking boundary conditions effectively narrows down possible solutions before checking permutations.

This method shows the strength of integer solutions: they allow flexibility within constraints while obeying strict operational parameters.

Linear Diophantine Equations

Linear Diophantine equations are equations in the form of \( ax + by = c \) where \(a\), \(b\), and \(c\) are integer constants, and solutions \(x\) and \(y\) are also integers. In this exercise, the equation \(4x + 6y = 100\) is one such Diophantine equation.

For a solution to exist, \(c\) (in this case, 100) must be divisible by the greatest common divisor of \(a\) (4) and \(b\) (6). Luckily here, 4 and 6 have a greatest common divisor of 2, and 100 is divisible by 2. So solutions are possible.

Solving Diophantine equations involves finding specific solutions using bounds and divisibility. The approach entails iterating potential solutions starting from simple and straightforward values to complex combinations, always ensuring the sum conditions (

• \(x + y + z = 20\)

) are met. This flexibility in solution finding highlights the power of linear Diophantine techniques in real-world problems, such as calculating cricket scores.