Problem 144 - A Complete Resource Book in Mathematics for JEE Main

Question

There are three pots and four coins. All these coins are to be distributed into these pots where any pot can contain any number of coins. In how many ways all these coins can be distributed if out of 4 coins 2 coins are identical and all pots are different? (A) 45 (B) 27 (C) 54 (D) None of these

Step-by-Step Solution

Answer

The answer is 54 (option C).

Step 1: Understanding the problem

We need to distribute 4 coins into 3 different pots, where 2 of the coins are identical. We need to count the number of distinct ways the coins can be distributed across the pots.

Step 2: Identical and distinct coin arrangements

Treat the problem as two sub-problems: one with the 2 identical coins and another with 2 distinct coins.

Step 3: Distribute identical coins

For the identical coins, each coin can be placed into any of the 3 pots. The number of ways to distribute both identical coins amongst the 3 pots is given by solving for non-negative integer solutions where the sum is equal to the number of coins. Using the formula \( x_1 + x_2 + x_3 = 2 \) where each \( x_i \) represents the number of identical coins in pot \( i \). The number of solutions is \( \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} = 6 \).

Step 4: Distribute distinct coins

Each of the two distinct coins can independently be placed into any of the 3 pots. For each distinct coin, there are 3 choices. Therefore, there are \( 3^2 = 9 \) ways to distribute the 2 distinct coins in the pots.

Step 5: Combine arrangements

To find the total number of ways to distribute all coins, multiply the number of arrangements for the identical coins with that of the distinct coins. This yields \( 6 \times 9 = 54 \).

Step 6: Select the correct answer

The number of ways all the coins can be distributed given the problem's conditions is 54, which matches option (C).

Key Concepts

Identical Items Distribution

When distributing identical items into containers, it's important to recognize that swapping identical items does not create a new distribution. In this problem, we have two identical coins that need to be placed into three different pots. This scenario is a classic example of partitioning where the order in which items are placed doesn't matter.

A helpful way to think about identical item distribution is by using the "stars and bars" theorem. This method helps in finding the number of non-negative integer solutions to the equation: \[ x_1 + x_2 + x_3 = 2 \] where each \( x_i \) represents the number of identical coins in pot \( i \). The formula \[ \binom{n + k - 1}{k - 1} \] where \( n \) is the total number of items and \( k \) is the number of groups, can be used to solve such problems. Here, \( n = 2 \) and \( k = 3 \), leading to \[ \binom{4}{2} = 6 \] ways.

Binomial Coefficient

The concept of the binomial coefficient is instrumental in calculating combinations, especially when dealing with identical items distribution or any other combinatorial scenarios. It provides a way to determine how many combinations can be formed when selecting items from a larger set.

In mathematical terms, the binomial coefficient, often represented as \[ \binom{n}{r} \] calculates the number of ways to choose \( r \) items from \( n \) items without considering the order. It is calculated using the formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where \(!\) denotes a factorial. This mathematical tool is key in the stars and bars method, which we used earlier to determine how to distribute two identical coins into three pots.

Disjoint Subsets

The mystical part about placing items into disjoint subsets is that each subset or group is unique, and items in different subsets do not interact or overlap. Imagine having distinct items, like our two distinct coins in the original exercise. Here, each coin can independently fall into any pot, forming subsets that don't affect each other.

To find the number of ways to place distinct coins into these pots, consider each choice separately. For each coin, there are three pot choices. Since there are two independent coins, we multiply the choices: \[ 3 \times 3 = 3^2 = 9 \] ways. Understanding disjoint subsets allows us to compartmentalize problems, dealing with each part separately while ensuring no overlap, as shown in the original problem's solution.