Problem 19
Question
Assuming \(k \leq n,\) in how many ways can we pass out \(k\) distinct pieces of fruit to \(n\) children if each child may get at most one? What is the number if \(k>n\) ? Assume for both questions that we pass out all the fruit.
Step-by-Step Solution
Verified Answer
For k ≤ n: \(\binom{n}{k} \times k!\). For k > n: 0.
1Step 1 - Understand the scenario
We need to distribute k distinct pieces of fruit to n children such that each child gets at most one piece.
2Step 2 - Case 1: When k ≤ n
Since each child can receive at most one piece, we are essentially selecting k out of n children to pass out the fruit. The number of ways to choose k children from n is given by the binomial coefficient \(\binom{n}{k}\). Once the children are chosen, we can distribute k distinct pieces of fruit among them in k! (k factorial) ways.
3Step 3 - Evaluate the formula for k ≤ n
The total number of ways is \(\binom{n}{k} \times k!\).
4Step 4 - Case 2: When k > n
If k > n, there are more pieces of fruit than children. It is impossible to distribute k distinct pieces with a limit of one piece per child. Thus, no distribution is possible in this case.
5Step 5 - Conclusion
For k ≤ n, the number of ways to distribute the fruit is \(\binom{n}{k} \times k!\). For k > n, the number of ways is 0.
Key Concepts
Binomial CoefficientFactorialCombinatorial Distribution
Binomial Coefficient
The binomial coefficient is a fundamental part of combinatorial mathematics. It provides a way to choose a specific number of elements from a larger set. We denote it as \(\binom{n}{k}\), where ‘n’ is the total number of elements and ‘k’ is the number of elements to be chosen. The formula for the binomial coefficient is:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Here, ‘!’ denotes a factorial, which we will explain in the next section.
For example, if you have 5 children and you want to choose 2 to receive fruit, you would calculate:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \]
This means there are 10 different ways to choose 2 children out of 5.
The binomial coefficient is essential in scenarios where you are selecting a subset of items or people, without considering the order.
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Here, ‘!’ denotes a factorial, which we will explain in the next section.
For example, if you have 5 children and you want to choose 2 to receive fruit, you would calculate:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \]
This means there are 10 different ways to choose 2 children out of 5.
The binomial coefficient is essential in scenarios where you are selecting a subset of items or people, without considering the order.
Factorial
A factorial, denoted by the symbol \(!\), is the product of all positive integers up to a specified number. For example, 5 factorial (written as 5!) is calculated as:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Factorials grow very quickly with larger numbers. They are crucial in permutations and combinations because they help calculate the number of ways to arrange or choose items.
In our problem, once we have selected children using the binomial coefficient, we need to distribute the distinct pieces of fruit. The distribution of k pieces of fruit among k children can occur in k! (k factorial) ways.
For instance, if you have 3 pieces of fruit and 3 children, you can distribute the fruits in:
\[ 3! = 3 \times 2 \times 1 = 6 \text{ ways} \]
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Factorials grow very quickly with larger numbers. They are crucial in permutations and combinations because they help calculate the number of ways to arrange or choose items.
In our problem, once we have selected children using the binomial coefficient, we need to distribute the distinct pieces of fruit. The distribution of k pieces of fruit among k children can occur in k! (k factorial) ways.
For instance, if you have 3 pieces of fruit and 3 children, you can distribute the fruits in:
\[ 3! = 3 \times 2 \times 1 = 6 \text{ ways} \]
Combinatorial Distribution
Combinatorial distribution focuses on how to distribute items among recipients under specific constraints. Our exercise deals with two primary cases.
If \(k ≤ n\), meaning the number of pieces of fruit (k) is less than or equal to the number of children (n), we use the following formula to find the number of possible distributions:
\[ \binom{n}{k} \times k! \]
This accounts for choosing k children from n and distributing the distinct pieces of fruit among them.
If \(k > n\), meaning the number of pieces of fruit exceeds the number of children, it is impossible to distribute each fruit piece to only one child. In this case, no distribution is possible, and the number of ways is 0.
Understanding these principles helps solve various real-world problems where limited resources need to be optimally allocated.
If \(k ≤ n\), meaning the number of pieces of fruit (k) is less than or equal to the number of children (n), we use the following formula to find the number of possible distributions:
\[ \binom{n}{k} \times k! \]
This accounts for choosing k children from n and distributing the distinct pieces of fruit among them.
If \(k > n\), meaning the number of pieces of fruit exceeds the number of children, it is impossible to distribute each fruit piece to only one child. In this case, no distribution is possible, and the number of ways is 0.
Understanding these principles helps solve various real-world problems where limited resources need to be optimally allocated.
Other exercises in this chapter
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