Problem 80 - Algebra for JEE ADVANCED

Question

Fifteen identical balls have to be put in five different boxes. Each box can contain any number of balls. The total number of ways of putting the balls into the boxes so that each box contains at least two balls is equal to a. \({ }^{9} C_{5}\) b. \({ }^{10} \mathrm{C}_{5}\) c. \({ }^{6} C_{5}\) d. \({ }^{10} C_{b}\)

Step-by-Step Solution

Answer

The answer is \(\binom{9}{4}\), so the correct option is a. \(^{9}C_{5}\).

Step 1: Understand the Problem

We need to distribute 15 indistinguishable balls into 5 distinguishable boxes, with each box having at least 2 balls. Our task is to find how many ways we can do so.

Step 2: Distribute Minimum Balls

Since each box must contain at least 2 balls, allocate 2 balls to each of the 5 boxes. This uses up 10 balls (i.e., \(2 \times 5 = 10\) balls).

Step 3: Calculate Remaining Balls

After placing 2 balls in each box, we have used 10 balls out of 15. Therefore, 5 balls remain to be distributed among the boxes freely.

Step 4: Apply Combinatorial Formula

The problem of distributing 5 identical balls into 5 boxes with no restrictions (since the minimum constraint has been met) is equivalent to finding the number of non-negative integer solutions to the equation \(x_1 + x_2 + x_3 + x_4 + x_5 = 5\).

Step 5: Use Stars and Bars Method

According to the stars and bars method, the number of solutions is given by \(\binom{n+k-1}{k-1}\), where \(n\) is the number of balls and \(k\) is the number of partitions (boxes). Here, \(n = 5\) and \(k = 5\), so we calculate \(\binom{5+5-1}{5-1} = \binom{9}{4}\).

Step 6: Calculate the Combination

Finally, compute \(\binom{9}{4}\) to find the number of ways: \[\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\].

Key Concepts

Stars and Bars Method

The stars and bars method is a very clever technique in combinatorics used to solve problems involving the distribution of indistinguishable objects (like balls) into distinguishable bins (like boxes). This method helps us count the ways to distribute these objects without worrying about the order in which they are placed.

Imagine you have a certain number of stars arranged in a line and you need to place dividers between them to form groups. Each arrangement of stars and bars corresponds to a unique distribution of the objects into bins. The number of dividers is always one less than the number of bins you have.

For example, given 5 balls to be placed in 5 boxes, the stars and bars method will visualize it as needing to place 4 dividers (bars) amongst the stars to separate them into 5 groups. The formula derived from this concept is \(\binom{n+k-1}{k-1}\), where \(n\) is the number of indistinguishable items, and \(k\) is the number of partitions. Thus, solving our problem involves calculating \(\binom{9}{4}\) to find the number of ways to distribute the remaining balls.

Binomial Coefficient

The binomial coefficient, often notated as \(\binom{n}{k}\), represents the number of ways to choose \(k\) elements from a set of \(n\) elements, regardless of order. It is a powerful tool in combinatorics for solving problems related to combinations and permutations, and it becomes especially handy in conjunction with the stars and bars method.

The binomial coefficient is calculated using the formula:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
where \(!\) denotes factorial, which is the product of all positive integers up to a given number. In our problem, we use the binomial coefficient \(\binom{9}{4}\), calculated as \(\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\). This tells us there are 126 ways to distribute 5 remaining balls into the 5 boxes.

Integer Solutions to Equations

Finding integer solutions to equations is common in combinatorics, especially when paired with the stars and bars method. We often need to solve equations like \(x_1 + x_2 + ... + x_k = n\), where \(x_i\) are non-negative integers representing how items are distributed among different categories or containers.

In our problem, after the initial allocation of 2 balls in each box, we need to solve the equation \(x_1 + x_2 + x_3 + x_4 + x_5 = 5\), with each \(x_i\) denoting the count of remaining balls in each box. Solving these equations allows us to count how many non-negative integer solutions are possible, using combinatorial tools like the binomial coefficient. This systematic approach ensures that all possible distributions are counted, providing us with the exact number of possible ways to solve the problem.