Problem 119
Question
In how many ways can 20 oranges be given to four children if each child should get at least one orange? (A) 869 (B) 969 (C) 973 (D) None of these
Step-by-Step Solution
Verified Answer
969 ways (option B).
1Step 1: Understand the Problem
We have 20 identical oranges and need to distribute them to four children. Each child must receive at least one orange.
2Step 2: Apply the Stars and Bars Method
Since each child must receive at least one orange, we will initially give each child one orange. This reduces the problem to distributing the remaining 16 oranges (because 20 - 4 = 16) to the four children.
3Step 3: Calculate Using Stars and Bars
With the problem now reduced to distributing 16 identical oranges to 4 children with no restrictions, we can use the formula for distributing \( n \) identical items into \( k \) groups: \[ \binom{n+k-1}{k-1} \]Here, \( n = 16 \) and \( k = 4 \), so the number of ways is:\[ \binom{16+4-1}{4-1} = \binom{19}{3} \]
4Step 4: Compute the Binomial Coefficient
Calculate \( \binom{19}{3} \):\[ \binom{19}{3} = \frac{19 \cdot 18 \cdot 17}{3 \cdot 2 \cdot 1} = 969 \]
5Step 5: Conclusion
The correct answer corresponds to option B: 969.
Key Concepts
Stars and Bars methodBinomial CoefficientDistribution of identical itemsProblem-solving in mathematics
Stars and Bars method
The Stars and Bars method is a powerful tool used in combinatorics to solve distribution problems. It helps us figure out how to allocate identical items, like oranges, among a certain number of groups or categories, such as children. The method is so named due to its visual representation, where stars represent the items to be distributed, and bars indicate the dividers between groups.
To apply this method, you first need to determine any initial constraints. In our problem, each child must get at least one orange. We start by giving each child one orange. This leaves us with 16 oranges to distribute freely.
To apply this method, you first need to determine any initial constraints. In our problem, each child must get at least one orange. We start by giving each child one orange. This leaves us with 16 oranges to distribute freely.
- Place the remaining items (stars) and dividers (bars) between them.
- For 16 items and 4 groups, it translates to finding positions for 3 dividers among the stars.
Binomial Coefficient
The binomial coefficient is defined as the number of ways you can choose a certain number of items from a set, regardless of the order. It's commonly denoted as \( \binom{n}{k} \), where \( n \) represents the total number of items, and \( k \) stands for the number of items to choose.
In the context of our exercise, the binomial coefficient allows us to calculate how many ways we can insert dividers among our stars. For distributing 16 identical items among 4 children, the required calculation is \( \binom{19}{3} \). This formula arises because we've altered the problem to one of distributing 16 stars between 3 dividers.
The actual value of \( \binom{19}{3} \) is determined by:
In the context of our exercise, the binomial coefficient allows us to calculate how many ways we can insert dividers among our stars. For distributing 16 identical items among 4 children, the required calculation is \( \binom{19}{3} \). This formula arises because we've altered the problem to one of distributing 16 stars between 3 dividers.
The actual value of \( \binom{19}{3} \) is determined by:
- Multiplying the numbers following 19, i.e., 19, 18, and 17.
- Then divide the result by 3 factorial (3 x 2 x 1) to consider only unique combinations.
Distribution of identical items
Distributing identical items among different groups is a common problem in combinatorics. Unlike distinct items, identical items pose unique challenges because the order doesn't matter, only the distribution among groups.
The key to solving these problems efficiently is recognizing patterns and applying combinatorial methods like Stars and Bars effectively. For instance, once we know each child must have at least one orange, we quickly shift to distributing remaining oranges without restrictions.
By using such techniques, we translate a complex problem into one that counts arrangements of stars and perceived boundaries (bars). This simplification aids in using mathematical tools to find the number of possible distributions, without manually listing all possibilities, which can be overwhelming with larger numbers.
The key to solving these problems efficiently is recognizing patterns and applying combinatorial methods like Stars and Bars effectively. For instance, once we know each child must have at least one orange, we quickly shift to distributing remaining oranges without restrictions.
By using such techniques, we translate a complex problem into one that counts arrangements of stars and perceived boundaries (bars). This simplification aids in using mathematical tools to find the number of possible distributions, without manually listing all possibilities, which can be overwhelming with larger numbers.
Problem-solving in mathematics
Approaching problem-solving in mathematics, especially combinatorics, requires clear understanding and methodical application. The steps involve not just reaching the answer but comprehending the process.
By understanding the process of using tools like Stars and Bars, and calculating binomial coefficients, students develop a robust framework for tackling a wide range of problems in mathematics, reinforcing both their problem-solving skills and their confidence.
- First, identify and clarify the problem statement.
- Look for ways to simplify complex scenarios without altering the core requirements.
By understanding the process of using tools like Stars and Bars, and calculating binomial coefficients, students develop a robust framework for tackling a wide range of problems in mathematics, reinforcing both their problem-solving skills and their confidence.
Other exercises in this chapter
Problem 117
The number of numbers greater than \(10^{6}\) that can be formed using the digits of the number 2334203, if all the digits of the given number must be used, is
View solution Problem 118
If ' \(n\) 'is an integer between 0 and 21 , then the minimum value of \(n !(21-n) !\) is (A) \(9 ! 2 !\) (B) \(10 ! 11 !\) (C) \(20 !\) (D) \(21 !\)
View solution Problem 120
The total number of 5 -digit numbers of different dig. its in which the digit in the middle is the largest is (A) \(\sum_{n=4}^{9}\left({ }^{n} P_{4}-{ }^{n-1}
View solution Problem 122
An \(n\)-digit number is a positive number with exactly \(n\) digits. Nine hundred distinct \(n\)-digit numbers are to be formed using only the three digits 2,5
View solution