Problem 161 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

There are two urns. Urn \(I\) has 3 distinct red balls and Urn II has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is (A) 36 (B) 66 (C) 108 (D) 3

Step-by-Step Solution

Answer

Option (C) 108 ways.

Step 1: Understand the Problem

We have two urns: Urn I with 3 distinct red balls and Urn II with 9 distinct blue balls. We need to transfer 2 balls from Urn I to Urn II and 2 balls from Urn II to Urn I, and find out how many distinct ways this can be done.

Step 2: Calculate Ways to Transfer Balls from Urn I to Urn II

Initially, we calculate how many ways we can choose 2 balls to transfer from Urn I to Urn II. Since Urn I has 3 distinct red balls, the number of ways to choose 2 balls out of 3 is given by the combination formula \( \binom{n}{k} \). Thus, we have \( \binom{3}{2} = 3 \) ways.

Step 3: Calculate Ways to Transfer Balls from Urn II to Urn I

Next, we calculate how many ways we can choose 2 balls to transfer from Urn II to Urn I. Since Urn II has 9 distinct blue balls, the number of ways to choose 2 balls out of 9 is \( \binom{9}{2} \). Thus, we have \( \binom{9}{2} = 36 \) ways.

Step 4: Calculate Total Ways to Transfer Balls

Since the transfers from Urn I to Urn II and from Urn II to Urn I are independent events, we multiply the number of ways calculated in Steps 2 and 3. So, the total number of ways is \( 3 \times 36 = 108 \).

Step 5: Choose the Correct Option

From the calculations, the total number of ways to transfer the balls is 108. Thus, the correct answer is option (C) 108.

Key Concepts

Understanding the Combination Formula

Combinations are a way to count the number of ways we can select a certain number of items from a larger set, without considering the order. This is perfect for situations where the arrangement does not matter, such as picking balls from an urn! The combination formula is used to calculate exactly how many ways we can make such selections.

The formula is given by:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

• \( n \) is the total number of items to choose from.
• \( k \) is the number of items to choose.
• \( ! \) denotes a factorial, which is the product of an integer and all the integers below it.

In the exercise, when choosing 2 balls out of 3 red ones in Urn I, we plug into the formula:\[\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3\]Similarly, for choosing 2 balls out of 9 blue ones in Urn II:\[\binom{9}{2} = \frac{9!}{2! \times 7!} = \frac{9 \times 8}{2 \times 1} = 36\]These calculated values help determine how many ways balls can be selected from the urns.

Independent Events Explained

In probability and combinatorics, events are independent when the outcome of one event does not influence the outcome of another. This is key to understanding why we multiply combinations in this problem.

When we calculate the ways to select balls from each urn separately, these are considered independent events. Picking balls from Urn I has no effect on the selections from Urn II, and vice versa. Therefore, we can determine the total number of outcomes by multiplying the number of ways to do each task separately.

For our problem:

• Number of ways to choose 2 out of 3 red balls from Urn I is 3 ways.
• Number of ways to choose 2 out of 9 blue balls from Urn II is 36 ways.

Because these events are independent, we multiply the two:\[3 \times 36 = 108\]This gives us the total number of ways to transfer the balls between urns.

Ball Selection Problem Insight

The ball selection problem presented here involves transferring sets of balls between two urns, and it's a common type of question in combinatorics.

The challenge is in understanding how to apply combinatorial concepts such as combinations and independent events to determine the number of outcomes. The specific scenario we analyzed involves the following steps:

• Select 2 red balls from a group of 3 in Urn I.
• Select 2 blue balls from a group of 9 in Urn II.
• Transfer them to the opposite urn.

It's crucial to break down the problem into pieces: first, consider each urn separately to determine possible selections, and then aggregate them using the principle of independent events.

By meticulously applying these concepts, we can confidently determine that there are 108 distinct ways to perform the ball transfer, according to the rules of the problem. This logical and methodical approach is often necessary for solving combinatorial tasks efficiently.