Question

find the number of ways in which 18 identical balls can be used in 15 different cricket matches.... find the number of ways in which 18 identical balls can be used in 15 different cricket matches such that at least one ball is used in one match.

Answer 1

680

Answer 2

The problem asks us to find the number of ways to distribute 18 identical balls into 15 different cricket matches such that at least one ball is used in one match.

Let $x_i$ be the number of balls used in the $i$-th cricket match, where $i$ ranges from 1 to 15. Since the balls are identical and the matches are different, we are looking for the number of integer solutions to the equation: $x_1 + x_2 + \dots + x_{15} = 18$

The phrase "at least one ball is used in one match" is crucial. This phrase can be interpreted in two ways:

1. Interpretation 1 (Common in combinatorics): At least one ball is used in each match. This means $x_i \ge 1$ for all $i=1, 2, \dots, 15$.
2. Interpretation 2 (Literal but often trivial): At least one ball is used in at least one match. This means that not all $x_i$ are zero. Since the total number of balls is 18, it is impossible for all $x_i$ to be zero (as their sum must be 18). Therefore, if this interpretation is taken, the condition is automatically satisfied, and the problem reduces to finding the number of non-negative integer solutions to the equation without any further restrictions, which would be $\binom{18+15-1}{15-1} = \binom{32}{14}$.

Given that a specific condition is added to the problem statement, it is highly probable that it is intended to impose a non-trivial restriction. In combinatorial problems involving distribution, "at least one" usually implies "at least one in each category". Thus, we proceed with Interpretation 1.

Interpretation 1: At least one ball is used in each match (i.e., $x_i \ge 1$ for all $i$).

To solve this, we can use the stars and bars method. Since each of the 15 matches must receive at least one ball, we first place one ball in each of the 15 matches. Number of balls used = $1 \times 15 = 15$. Number of balls remaining = $18 - 15 = 3$.

Now, we need to distribute these 3 remaining identical balls among the 15 different matches. There are no further restrictions on these remaining balls (a match can receive zero or more of these additional balls). Let $y_i$ be the number of additional balls given to match $i$. Then, we need to find the number of non-negative integer solutions to the equation: $y_1 + y_2 + \dots + y_{15} = 3$

Here, $n = 3$ (the number of remaining identical items, i.e., stars) and $k = 15$ (the number of distinct bins, i.e., bars needed is $k-1$). The number of non-negative integer solutions is given by the stars and bars formula: $\binom{n+k-1}{k-1}$ or $\binom{n+k-1}{n}$.

Substituting $n=3$ and $k=15$: Number of ways = $\binom{3 + 15 - 1}{15 - 1}$ Number of ways = $\binom{17}{14}$

Alternatively, using the other form: Number of ways = $\binom{3 + 15 - 1}{3}$ Number of ways = $\binom{17}{3}$

Both expressions are equivalent, as $\binom{N}{R} = \binom{N}{N-R}$. Let's calculate the value of $\binom{17}{3}$: $\binom{17}{3} = \frac{17!}{3!(17-3)!} = \frac{17!}{3!14!} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1}$ $\binom{17}{3} = 17 \times \frac{16}{2} \times \frac{15}{3}$ $\binom{17}{3} = 17 \times 8 \times 5$ $\binom{17}{3} = 17 \times 40$ $\binom{17}{3} = 680$

The number of ways is 680.

The final answer is $\boxed{680}$.

Explanation of the solution:

The problem requires distributing 18 identical balls into 15 different matches such that each match receives at least one ball.

1. Assign one ball to each of the 15 matches. This uses 15 balls.
2. Remaining balls: $18 - 15 = 3$.
3. Distribute these 3 remaining identical balls among the 15 different matches (any match can receive 0 or more).
4. This is a stars and bars problem with $n=3$ (stars) and $k=15$ (bins).
5. The number of ways is $\binom{n+k-1}{k-1} = \binom{3+15-1}{15-1} = \binom{17}{14}$.
6. $\binom{17}{14} = \binom{17}{3} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = 17 \times 8 \times 5 = 680$.