Problem 162 - A Complete Resource Book in Mathematics for JEE Main

Question

Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is \({ }^{9} C_{3}\) Statement-2: The number of ways of choosing any 3 places from 9 different places is \({ }^{9} C_{3}\). (A) Statement- 1 is true, Statement- 2 is true; Statement- 2 is not a correct explanation for Statement- 1 (B) Statement- 1 is true, Statement- 2 is false. (C) Statement- 1 is false, Statement- 2 is true. (D) Statement- 1 is true, Statement- 2 is true; Statement-2 is a correct explanation for Statement- 1

Step-by-Step Solution

Answer

(C) Statement-1 is false, Statement-2 is true.

Step 1: Understanding Statement-1

Statement-1 claims that the number of ways of distributing 10 identical balls into 4 distinct boxes such that no box is empty is given by \( {^9}C_3 \). To solve this, we can use the "stars and bars" method for the distribution of identical items into distinct groups with each group receiving at least one item. We first give 1 ball to each box to ensure they are not empty. This uses up 4 balls, leaving us with 6 balls to distribute freely among the 4 boxes. The formula for distributing \(n\) identical items into \(k\) distinct boxes with no empty box allowed is \( {^{n-1}}C_{k-1} \). Here, \(n=6\) and \(k=4\), so \( {^{n-1}}C_{k-1} = {^5}C_3 \), not \( {^9}C_3 \), hence Statement-1 is false.

Step 2: Understanding Statement-2

Statement-2 states that the number of ways of choosing any 3 places from 9 different places is \( {^9}C_3 \). The combination formula \( {^n}C_r \) expresses the number of ways to choose \(r\) items from a set of \(n\) items without regard to order. Substituting \(n=9\) and \(r=3\) into this formula gives \( {^9}C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \). Therefore, Statement-2 is true.

Step 3: Conclusion on Statements

Statement-1 is false because the correct computation for distributing balls using the stars and bars method does not result in \( {^9}C_3 \). Statement-2 is true as it accurately describes a valid combination problem and its solution. Hence, the correct option is (C), which states that Statement-1 is false and Statement-2 is true.

Key Concepts

Stars and Bars Method

The concept of the "Stars and Bars" method might sound a bit abstract at first, but it's a powerful tool in combinatorics for distributing identical items into distinct groups. Imagine you have a set number of items, in this case, identical balls, that you want to distribute among several distinct boxes. The challenge is to compute how many different ways this distribution can occur.

Here's where the stars (representing items) come into play, and the bars (representing divisions between groups) help in organizing these items. A key point when using this technique is to make sure each box gets at least one item to start with. Once we ensure each box is non-empty, the task is to distribute the remaining items.

• Step 1: Assign one item to each box to ensure they are not empty. This directly reduces the number of items to be distributed freely.
• Step 2: Use the formula \( n - 1 \) choose \( k - 1 \), where \( n \) is the remaining number of items and \( k \) is the number of boxes.

For instance, if you began with 10 identical items and 4 boxes, giving one item to each box leaves you with 6 items to distribute. The number of distributions, while ensuring no box is empty, would be \( {^{5}}C_{3} \), due to 6-1 \( = 5\) items and 4-1 \( = 3\) divisions.

Permutations and Combinations

In combinatorics, understanding the difference between permutations and combinations is crucial. Permutations concern the arrangement of objects where order matters, whereas combinations focus on the selection of objects where order doesn't matter.

Printed on any combination lock are numbers - these represent a classic problem where only the combination matters, not the order. In mathematical notation, combinations are denoted by \( {^n}C_r \), where \( n \) is the total number of available items and \( r \) is the number of items to choose.

• Formula for Combinations: \( {^n}C_r = \frac{n!}{r!(n-r)!} \)
• Order is NOT important in combinations.

Returning to our exercise, the task was to find out how many ways we can choose 3 places from a set of 9. The solution, \( {^9}C_3 \), describes how combinations work: selecting 3 items (positions) without consideration of the order in which they are chosen, yielding 84 possible "combinations." This shows just how combinations are efficient for tasks where only the end group matters, not the sequence.

Distribution of Identical Items

Distributing identical items into distinct groups, like in the original exercise, poses a unique challenge. Here, the key is recognizing that the items are indistinct and identical. This has considerable implications because every arrangement that meets the counting constraints is considered identical.

The idea of distributing identical items can also be aligned with the concept of "partitioning" numbers, where you split one number into a sum of smaller numbers (representing distributed items) with specific conditions (e.g., each box gets at least one item).

• The constraint can usually be that no group remains empty, as in our box problem.
• The "stars and bars" method offers an excellent systematic solution to this issue.

Whenever dealing with identical items, those distributed quantities can't be distinctly ordered among themselves; hence, the solution considers how partitions with constraints can be calculated, differing from the permutation approach which focuses on ordering distinct items.