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Chapter 1: Problem 37

In how many ways can we pass out \(k\) (identical) ping-pong balls to \(n\) children if each child may get at most one? (h)

Short Answer

Expert verified
\(\binom{n}{k}\)

Step by step solution

01

Understand the problem

The problem requires distributing a maximum of 1 ping-pong ball to each child such that each child gets at most one ball and there are a total of k ping-pong balls and n children.
02

Consider the conditions

Each child can either receive 1 ping-pong ball or none. The total number of ways to distribute these balls has to satisfy the condition that no child gets more than one ball.
03

Determine the number of children and balls

Identify the number of children, n, and the number of ping-pong balls, k.
04

Apply combinations formula

Since each child can get a maximum of one ball, the problem boils down to selecting k children out of n to give the balls. The number of ways to choose k children out of n is given by the binomial coefficient: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Coefficient
The binomial coefficient is a key concept in combinatorics. It represents the number of ways to choose a subset of items from a larger set, without considering the order of the items. Mathematically, it is denoted as \(\binom{n}{k}\), and it reads as 'n choose k'. This is computed using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(!\) denotes the factorial operator. The binomial coefficient tells you how many different ways you can select \(k\) balls from \(n\) children, assuming each child can receive at most one ball. It simplifies the distribution problem and makes it manageable using straightforward arithmetic calculations.
Combinatorial Distribution
Combinatorial distribution deals with the strategies to distribute items among groups following specific rules. In our problem, we are distributing \(k\) identical ping-pong balls to \(n\) children, each child getting at most one ball. The key here is that once the positions for the balls are chosen (which children will receive the balls), the actual distribution is straightforward because each child can get only one ball or none. This means we need to count the combinations rather than permutations. The binomial coefficient \(\binom{n}{k}\) gives us the exact number of ways to accomplish this distribution.
Factorials
Factorials play a significant role in combinatorics and are crucial to understanding how to calculate the binomial coefficient. The factorial of a positive integer \(n\), denoted as \(n! \), is the product of all positive integers less than or equal to \(n\). For example, \(!5 = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow very quickly with increasing \(n\), so while the calculations can become large, they simplify expressions in combinatorial formulas. When you compute \(\binom{n}{k}\), factorials help manage the different ways of choosing subsets, ensuring both the numerator and denominator account for all the possible selections and arrangements.

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Most popular questions from this chapter

Let \(C\) be the set of \(k\) -element subsets of \([n]\) that contain the number \(n,\) and let \(D\) be the set of \(k\) -element subsets of \([n]\) that don't contain \(n .\) (a) Let \(C^{\prime}\) be the set of \((k-1)\) -element subsets of \([n-1] .\) Describe a bijection from \(C\) to \(C^{\prime}\). (A verbal description is fine.) (b) Let \(D^{\prime}\) be the set of \(k\) -element subsets of \([n-1]=\\{1,2, \ldots n-1\\}\). Describe a bijection from \(D\) to \(D^{\prime} .\) (A verbal description is fine.) (c) Based on the two previous parts, express the sizes of \(C\) and \(D\) in terms of binomial coefficients involving \(n-1\) instead of \(n\). (d) Apply the sum principle to \(C\) and \(D\) and obtain a formula that expresses \(\left(\begin{array}{l}n \\ k\end{array}\right)\) in terms of two binomial coefficients involving \(n-1\). You have just derived the Pascal Equation that is the basis for the famous Pascal's Triangle.

Now some number \(n\) of schools are going to send their baseball teams to a tournament, and each team must play each other team exactly once. Let us think of the teams as numbered 1 through \(n\). (a) How many games does team 1 have to play in? (b) How many games, other than the one with team \(1,\) does team two have to play in? (c) How many games, other than those with the first \(i-1\) teams, does team \(i\) have to play in? (d) In terms of your answers to the previous parts of this problem, what is the total number of games that must be played?

A tennis club has \(2 n\) members. We want to pair up the members by twos for singles matches. (a) In how many ways may we pair up all the members of the club? (Hint: consider the cases of \(2,4,\) and 6 members.) (h) (b) Suppose that in addition to specifying who plays whom, for each pairing we say who serves first. Now in how many ways may we specify our pairs?

Another kind of geometric path in the plane is a diagonal lattice path. Such a path is a path made up of line segments that go from a point \((i, j)\) to \((i+1, j+1)\) (this is often called an upstep) or \((i+1, j-1)\) (this is often called a downstep), again where \(i\) and \(j\) are integers. (Thus diagonal lattice paths always move towards the right but may move up or down.) (a) Describe which points are connected to (0,0) by diagonal lattice paths. (h) (b) What is the length of a diagonal lattice path from (0,0) to \((m, n) ?\) (c) Assuming that \((m, n)\) is such a point, how many diagonal lattice paths are there from (0,0) to \((m, n) ?(\mathrm{~h})\)

The word permutation is actually used in two different ways in mathematics. A permutation of a set \(S\) is one-to-one from \(S\) onto \(S\). How many permutations does an \(n\) -element set have?
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