Problem 73 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(n\) and \(k\) be positive integers such that \(n \geq \frac{k(k+1)}{2}\) The number of solutions \(\left(x_{1}, x_{2}, \ldots, x_{k}\right), x_{1} \geq 1, x_{2} \geq 2\), \(\ldots, x_{\mathrm{k}} \geq k\), all integers, satisfying \(x_{1}+x_{2}+\ldots+x_{\mathrm{k}}=n\), is (A) \({ }^{\mathrm{m}} \mathrm{C}_{\mathrm{k}-1}\) (B) \({ }^{\mathrm{m}} \mathrm{C}_{\mathrm{k}}\) (C) \({ }^{\mathrm{m}} \mathrm{C}_{\mathrm{k}+1}\) (D) none of these where \(m=\frac{1}{2}\left(2 n-k^{2}+k-2\right)\)

Step-by-Step Solution

Answer

(A) \(^mC_{k-1}\) is the correct answer.

Step 1: Understand the problem constraints

We need to find how many solutions exist for the equation \(x_1 + x_2 + \ldots + x_k = n\) with the conditions \(x_1 \geq 1\), \(x_2 \geq 2\), ..., \(x_k \geq k\).

Step 2: Introduce new variables

Let \(y_1 = x_1 - 1\), \(y_2 = x_2 - 2\), ..., \(y_k = x_k - k\). The new variables \(y_i\) satisfy \(y_i \geq 0\). The equation becomes \(y_1 + 1 + y_2 + 2 + \ldots + y_k + k = n\).

Step 3: Simplify the equation

We substitute the new variables and simplify: \(y_1 + y_2 + \ldots + y_k + \frac{k(k+1)}{2} = n\). This leads to \(y_1 + y_2 + \ldots + y_k = n - \frac{k(k+1)}{2}\).

Step 4: Calculate the transformed sum conditions

The condition \(y_1 + y_2 + \ldots + y_k = m\) where \(m = n - \frac{k(k+1)}{2}\). This is transformed into a problem of finding the number of solutions in nonnegative integers \(y_1, y_2, \ldots, y_k\) such that their sum is \(m\).

Step 5: Apply stars and bars

Using the stars and bars method, the number of solutions is given by the combinatorial formula \(\binom{m+k-1}{k-1}\).

Step 6: Calculate \(m+k-1\) in terms of \(m\) and verify with options

We have \(m = n - \frac{k(k+1)}{2}\). Substituting for \(m\) in option formula gives us:\(m + k - 1 = \frac{1}{2}(2n - k^2 + k - 2) + k - 1\). Therefore, the combination we need to find is \(\binom{m+k-1}{k-1}\) matching \(\binom{\left(\frac{1}{2}(2n - k^2 + k -2) + k - 1\right)}{k-1}\), which matches option (A) \(\binom{m}{k-1}\).

Step 7: Conclusion

Based on the above calculations, the correct option is (A): \(^mC_{k-1}\) given \(m = \frac{1}{2}(2n - k^2 + k - 2)\).

Key Concepts

Stars and Bars Method

The "stars and bars method" is a classic combinatorial technique used to determine how many ways you can distribute indistinguishable items ("stars") into distinguishable bins ("bars"). This method is particularly useful for solving problems where we need to find all possible combinations of positive integer solutions to equations. Think of the stars as representing parts of a total to be divided among groups. The bars act as dividers between the groups. For example, if you want to divide 10 candies (stars) among 3 children (bars), you can imagine dividing a row of 10 stars by placing 2 bars between them. The total number of ways to arrange the stars and bars can be calculated using the binomial coefficient formula: \[ \binom{n+k-1}{k-1} \]where \( n \) is the total number of stars, and \( k \) is the number of bars (minus one, as you need one less bar than the number of groups).This concept is elegantly used in the given problem to solve equations with specific restrictions on integer solutions.

Integer Equations

In combinatorics, solving integer equations often involves finding the number of ways to express an integer \( n \) as the sum of other integers under given conditions. These conditions specify whether the integers are non-negative, positive, or have other constraints like being greater than or equal to another integer. The challenge lies in calculating how many different sets of integers satisfy the equation. In our problem, it's about finding solutions for an equation of the form \( x_1 + x_2 + \ldots + x_k = n \) with each integer \( x_i \) meeting the criteria of being greater or equal to values like 1, 2, and so forth up to \( k \). This turns into a problem of calculating non-negative integer solutions after adjusting variables through suitable transformations.

Positive Integers

Positive integers are a set of all natural numbers greater than zero. They are fundamental in solving equations where each component must be larger than a specified minimum threshold. For example, in the provided exercise, each \( x_i \) must be at least \( i \), leading to adjustments where \( x_1 \geq 1, x_2 \geq 2, ..., x_k \geq k \). Such constraints require transforming the problem into one solvable using the stars and bars method, focusing on non-negative integer solutions after compensating for these initial values.This is achieved by redefining variables as introduced in the step-by-step solution, ensuring we handle each piece as a non-negative integer part of a simpler equation.

Combinatorial Formulas

Combinatorial formulas are mathematical tools that help solve problems related to counting and arrangement. They simplify determining the number of ways to choose subsets, arrange elements, or partition sets into groups. One such formula extensively used is the binomial coefficient, represented as \( \binom{n}{k} \), which denotes the number of ways to choose \( k \) elements from \( n \) elements without considering order.In the context of the exercise, after adjusting variable constraints, we apply combinatorial formulas to find the number of non-negative integer solutions. Specifically, we used:\[ \binom{m+k-1}{k-1} \]This particular formula is derived from the stars and bars method and is crucial for determining the exact count of compositions of an integer into distinct parts for given conditions, as calculated in our exercise to ascertain the correct number of solutions.