Introduction to Algorithms
Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter C, Problem 1P

(a)

Program Plan Intro

To argue that numbers of ways of placing the balls in bins is bn .

(a)

Expert Solution
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Explanation of Solution

Given information:

The n balls are distinct and their order within bin doesn’t matter.

Explanation:

There can be b different decisions made for n balls about their placement. The total number of possibilities is just bn as the number of possible decisions about the n balls placement is independent of previous choices.

(b)

Program Plan Intro

To prove that there are exactly (b+n1)!(b1)! ways to place the balls in the bins.

(b)

Expert Solution
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Explanation of Solution

Given information:

It is assumed that balls are distinct and that balls in each bin are ordered.

Explanation:

First assume that sticks can be distinguished. This implies that there are total of n balls and b1 sticks to be arranged. This makes total things to be arranged as n+b1 and number of ways to do this is (b+n1)! . However some of these arrangements will be same and no matter how b1 sticks are arranged the answer will remain same. Thus n distinct balls and b1 indistinct sticks can be arranged in (n+b1)!(b1)! ways.

This arrangement can be related with the original statement, where sticks can be imagined as dividing lines between bins and ordered balls between them can be imagined as ordered balls in each bin.

(c)

Program Plan Intro

To show that (b+n1n) represents the numbers of ways of placing the balls in the bins.

(c)

Expert Solution
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Explanation of Solution

Given information:

The balls are identical and their order within a bin does not matter.

Explanation:

Using results from above two parts, it can be noticed that any of the n permutation of balls will result in the similar configuration. Thus, count from the previous parts must be divided by n! . Thus number of permutations left are (n+b1)!n!(b1)!=(n+b1n) .

(d)

Program Plan Intro

To show that number of ways of placing the balls is (bn) based on condition that nb .

(d)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The balls are identical and no bin may contain more than one ball.

Explanation:

Here, a set of bins to contain balls is selected,as each bin can have a ball or not. The numbers of bins selected is n since number of non-empty bins and the numbers of balls must be equal. In other words, a subset of size n of the bins is being selected from the whole set of bins. This becomes the combinatorial definition of (bn) defined in terms of selecting subsets.

(e)

Program Plan Intro

To show that number of ways of placing the balls is (n1b1) based on condition that nb .

(e)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The balls are identical and no bin may be left empty.

Explanation:

The condition is to put one ball in each bin as no bin can be left empty. Thus there are nb balls left. Now since each bin has at least one ball, so balls can be put into the bins with no further restriction. This is similar to case shown in part c. The main difference being that the number of balls to be distributed is nb rather than n. Thus the answer becomes ((nb)+b1nb)=(n1nb)=(n1(n1)(nb))=(n1b1) .

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