Problem 91 - Algebra for JEE ADVANCED

Question

The total number of ways in which \(n^{2}\) number of identical balls can be put in \(n\) numbered boxes \((1,2,3, \ldots, n)\) such that \(i^{\text {it }}\) box contains at least \(i\) number of balls is a. \({ }^{n^{2}} C_{s-1}\) b. \(\quad{ }^{n^{2}-1} C_{n-1}\) c. \(\quad \frac{\mu^{2}+u-2}{2} C_{H-1}\) d. none of these

Step-by-Step Solution

Answer

The answer is (c) \( \frac{n^2+u-2}{2} C_{H-1} \).

Step 1: Understand the Problem

We need to find the number of ways to distribute \(n^2\) identical balls into \(n\) boxes such that box \(i\) contains at least \(i\) balls.

Step 2: Calculate Minimum Balls Requirement

Each box \(i\) must contain at least \(i\) balls. Hence, box 1 must contain at least 1 ball, box 2 at least 2 balls, and so on. The sum of these minimum balls is given by \(1 + 2 + 3 + \, \ldots \, + n\). This sum is \(\frac{n(n+1)}{2}\).

Step 3: Determine Remaining Balls

Subtract the sum of minimum required balls from the total number of balls \(n^2\) to find the remaining balls to distribute freely. This is given by \(n^2 - \frac{n(n+1)}{2}\).

Step 4: Simplify Expression for Remaining Balls

Simplify the expression: \[ n^2 - \frac{n(n+1)}{2} = \frac{2n^2 - n(n+1)}{2} = \frac{2n^2 - n^2 - n}{2} = \frac{n^2 - n}{2} \]

Step 5: Distribute Remaining Balls Freely

Now, we need to distribute \(\frac{n^2 - n}{2}\) identical balls into \(n\) boxes. The formula for distributing \(k\) identical items into \(n\) boxes with no restrictions is given by \( \binom{k+n-1}{n-1} \).

Step 6: Apply Formula

Use the formula with \(k = \frac{n^2 - n}{2}\), giving us: \[ \binom{ \frac{n^2 - n}{2} + n - 1}{n - 1} = \binom{\frac{n^2 + n - 2}{2}}{n - 1} \]

Step 7: Match with Options

Compare the above expression with the given options. Note that option \(c\) matches \(\frac{n^2+n-2}{2}C_{n-1}\).

Key Concepts

Identical Objects Distribution

Distributing identical objects, like balls, into distinct groups, such as boxes, is a classic problem in combinatorics. It's crucial in understanding various counting principles. Here, all balls are identical, meaning each ball is indistinguishable from another. When asked to distribute these identical items, the challenge is to determine how combinations can be made if given targets or restrictions. Consider that each box can hold any number of balls with the only limit being the total number available. For easier understanding, relate it to dividing up resources or tasks where changing the assignment doesn't make the outcome different, only the count in each bucket matters.

Binomial Coefficient

Binomial coefficients are fundamental in combinatorics. They show how to choose items from a group, regardless of the sequence.In mathematical terms, we represent them as \( \binom{n}{k} \), denoting the ways to select \(k\) items from \(n\) total. This becomes particularly useful when calculating combinations where the order doesn't matter, as in our exercise.Think of it as distributing candies among children indiscriminately or dropping identical balls into different buckets. The binomial coefficient applies when these items move about without concerns for sequence.

Restriction in Distribution

Restrictions add complexity but also clarity to distribution problems. Typically, these restrictions clarify minimums or maximums regarding what each group can hold.In this exercise, every box must initially contain a set number of balls corresponding to its number, like box 1 having at least 1 ball, and so forth. This is calculated using the sequence sum: \[ 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \]This formula determines the non-flexible part of distribution which must be adhered to first. Restrictions guide the initial setup, ensuring each group's fundamental needs are met before distributing any surplus freely.

Algebraic Manipulation

Algebraic manipulation involves reworking equations to simplify or solve them, allowing us to see relationships and results more plainly. It’s a common tool in combinatorics to refine complex expressions.In our problem, we started with a subtraction \( n^2 - \frac{n(n+1)}{2} \) to find leftover balls. Through breakdown: \[ n^2 - \frac{n(n+1)}{2} = \frac{n^2 - n}{2} \] It becomes clear how many balls are still available for open allocation. Simplification helps in exploring further distributions without dealing with too overladen expressions, revealing the heart of what needs to be calculated next—the free distribution.