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

Question

Want to see more full solutions like this?

Answer 1

Question

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

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+n−1)!(b−1)! ways to place the balls in the bins.

(b)

Expert Solution

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 b−1 sticks to be arranged. This makes total things to be arranged as n+b−1 and number of ways to do this is (b+n−1)! . However some of these arrangements will be same and no matter how b−1 sticks are arranged the answer will remain same. Thus n distinct balls and b−1 indistinct sticks can be arranged in (n+b−1)!(b−1)! 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+n−1n) represents the numbers of ways of placing the balls in the bins.

(c)

Expert Solution

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+b−1)!n!(b−1)!=(n+b−1n) .

(d)

Program Plan Intro

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

(d)

Expert Solution

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 (n−1b−1) based on condition that n≥b .

(e)

Expert Solution

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 n−b 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 n−b rather than n. Thus the answer becomes ((n−b)+b−1n−b)=(n−1n−b)=(n−1(n−1)−(n−b))=(n−1b−1) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

CSE330 Discrete Mathematics 1. In the classes, we discussed three forms of floating number representations as given below, (1) Standard/General Form, (2) Normalized Form, (3) Denormalized Form. 3. Consider the real number x = (3.395) 10 (a) (b) Convert the decimal number x into binary format up to 7 binary places (7 binary digits after decimal) Convert the calculated value into denormalized form and calculate fl(x) for m=4 Don't use any Al tool show answer in pen a nd paper then take pi ctures and send

Simplify the following expressions by means of a four-variable K-Map. AD+BD+ BC + ABD

CSE330 Discrete Mathematics 1. In the classes, we discussed three forms of floating number representations as given below, (1) Standard/General Form, (2) Normalized Form, (3) Denormalized Form. 2. Let ẞ 2, m = 6, emin = -3 and emax = 3. Answer the following questions: Compute the minimum of |x| for General and Normalized form (a) Compute the Machine Epsilon value for the General and Denormalized form. If we change the value of emax to 6 then how will it affect the value of maximum scale invariant error for the case of Normalized form? Explain your answer. show answer in pen a Don't use any Al tool nd paper then take pi ctures and send

Answer 2

Question

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

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+n−1)!(b−1)! ways to place the balls in the bins.

(b)

Expert Solution

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 b−1 sticks to be arranged. This makes total things to be arranged as n+b−1 and number of ways to do this is (b+n−1)! . However some of these arrangements will be same and no matter how b−1 sticks are arranged the answer will remain same. Thus n distinct balls and b−1 indistinct sticks can be arranged in (n+b−1)!(b−1)! 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+n−1n) represents the numbers of ways of placing the balls in the bins.

(c)

Expert Solution

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+b−1)!n!(b−1)!=(n+b−1n) .

(d)

Program Plan Intro

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

(d)

Expert Solution

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 (n−1b−1) based on condition that n≥b .

(e)

Expert Solution

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 n−b 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 n−b rather than n. Thus the answer becomes ((n−b)+b−1n−b)=(n−1n−b)=(n−1(n−1)−(n−b))=(n−1b−1) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

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

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+n−1)!(b−1)! ways to place the balls in the bins.

(b)

Expert Solution

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 b−1 sticks to be arranged. This makes total things to be arranged as n+b−1 and number of ways to do this is (b+n−1)! . However some of these arrangements will be same and no matter how b−1 sticks are arranged the answer will remain same. Thus n distinct balls and b−1 indistinct sticks can be arranged in (n+b−1)!(b−1)! 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+n−1n) represents the numbers of ways of placing the balls in the bins.

(c)

Expert Solution

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+b−1)!n!(b−1)!=(n+b−1n) .

(d)

Program Plan Intro

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

(d)

Expert Solution

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 (n−1b−1) based on condition that n≥b .

(e)

Expert Solution

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 n−b 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 n−b rather than n. Thus the answer becomes ((n−b)+b−1n−b)=(n−1n−b)=(n−1(n−1)−(n−b))=(n−1b−1) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

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

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+n−1)!(b−1)! ways to place the balls in the bins.

(b)

Expert Solution

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 b−1 sticks to be arranged. This makes total things to be arranged as n+b−1 and number of ways to do this is (b+n−1)! . However some of these arrangements will be same and no matter how b−1 sticks are arranged the answer will remain same. Thus n distinct balls and b−1 indistinct sticks can be arranged in (n+b−1)!(b−1)! 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+n−1n) represents the numbers of ways of placing the balls in the bins.

(c)

Expert Solution

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+b−1)!n!(b−1)!=(n+b−1n) .

(d)

Program Plan Intro

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

(d)

Expert Solution

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 (n−1b−1) based on condition that n≥b .

(e)

Expert Solution

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 n−b 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 n−b rather than n. Thus the answer becomes ((n−b)+b−1n−b)=(n−1n−b)=(n−1(n−1)−(n−b))=(n−1b−1) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

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

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+n−1)!(b−1)! ways to place the balls in the bins.

(b)

Expert Solution

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 b−1 sticks to be arranged. This makes total things to be arranged as n+b−1 and number of ways to do this is (b+n−1)! . However some of these arrangements will be same and no matter how b−1 sticks are arranged the answer will remain same. Thus n distinct balls and b−1 indistinct sticks can be arranged in (n+b−1)!(b−1)! 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+n−1n) represents the numbers of ways of placing the balls in the bins.

(c)

Expert Solution

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+b−1)!n!(b−1)!=(n+b−1n) .

(d)

Program Plan Intro

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

(d)

Expert Solution

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 (n−1b−1) based on condition that n≥b .

(e)

Expert Solution

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 n−b 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 n−b rather than n. Thus the answer becomes ((n−b)+b−1n−b)=(n−1n−b)=(n−1(n−1)−(n−b))=(n−1b−1) .

Answer 6

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

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.

Answer 7

Chapter C, Problem 1P

(a)

Program Plan Intro

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

Answer 8

(a)

Expert Solution

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.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

To prove that there are exactly (b+n−1)!(b−1)! ways to place the balls in the bins.

(b)

Expert Solution

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 b−1 sticks to be arranged. This makes total things to be arranged as n+b−1 and number of ways to do this is (b+n−1)! . However some of these arrangements will be same and no matter how b−1 sticks are arranged the answer will remain same. Thus n distinct balls and b−1 indistinct sticks can be arranged in (n+b−1)!(b−1)! 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.

Answer 13

(b)

Program Plan Intro

To prove that there are exactly (b+n−1)!(b−1)! ways to place the balls in the bins.

Answer 14

(b)

Expert Solution

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 b−1 sticks to be arranged. This makes total things to be arranged as n+b−1 and number of ways to do this is (b+n−1)! . However some of these arrangements will be same and no matter how b−1 sticks are arranged the answer will remain same. Thus n distinct balls and b−1 indistinct sticks can be arranged in (n+b−1)!(b−1)! 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.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Program Plan Intro

To show that (b+n−1n) represents the numbers of ways of placing the balls in the bins.

(c)

Expert Solution

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+b−1)!n!(b−1)!=(n+b−1n) .

Answer 19

(c)

Program Plan Intro

To show that (b+n−1n) represents the numbers of ways of placing the balls in the bins.

Answer 20

(c)

Expert Solution

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+b−1)!n!(b−1)!=(n+b−1n) .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Program Plan Intro

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

(d)

Expert Solution

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.

Answer 25

(d)

Program Plan Intro

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

Answer 26

(d)

Expert Solution

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.

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

(e)

Program Plan Intro

To show that number of ways of placing the balls is (n−1b−1) based on condition that n≥b .

(e)

Expert Solution

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 n−b 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 n−b rather than n. Thus the answer becomes ((n−b)+b−1n−b)=(n−1n−b)=(n−1(n−1)−(n−b))=(n−1b−1) .

Answer 31

(e)

Program Plan Intro

To show that number of ways of placing the balls is (n−1b−1) based on condition that n≥b .

Answer 32

(e)

Expert Solution

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 n−b 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 n−b rather than n. Thus the answer becomes ((n−b)+b−1n−b)=(n−1n−b)=(n−1(n−1)−(n−b))=(n−1b−1) .