Number of ways to distribute 20 identical pencils to 6 non-identical children with restrictions The Next CEO of Stack OverflowNumber of ways distribute 12 identical action figures to 5 childrenProbability and Statistics (Normal Distribution)IN how many ways we can distribute 10 identical looking pencils given the following conditions?Determine number of ways to distribute sixteen identical pieces of candy to five non-identical kids…What is the difference between ways to distribute identical and different objects?Identical and different object distribution among childrenCombinatorics question involving distributing identical candies to different childrenCombinatorics question involving distributing 9 different candies to three different kidsMuddy Children Puzzle (when more than 2 kids are muddy)In how many ways can 9 different books and 17 identical cakes be divided on 6 children, so that each child receive at least one book and one cake?
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Number of ways to distribute 20 identical pencils to 6 non-identical children with restrictions
The Next CEO of Stack OverflowNumber of ways distribute 12 identical action figures to 5 childrenProbability and Statistics (Normal Distribution)IN how many ways we can distribute 10 identical looking pencils given the following conditions?Determine number of ways to distribute sixteen identical pieces of candy to five non-identical kids…What is the difference between ways to distribute identical and different objects?Identical and different object distribution among childrenCombinatorics question involving distributing identical candies to different childrenCombinatorics question involving distributing 9 different candies to three different kidsMuddy Children Puzzle (when more than 2 kids are muddy)In how many ways can 9 different books and 17 identical cakes be divided on 6 children, so that each child receive at least one book and one cake?
$begingroup$
How many ways are there to pass out 20 pencils (assume all the pencils are identical the same) to six children?
Based on the following condition:
a) No restriction. ( i.e. each kid may receive zero to 20 pencils.)
b) Every child receives at least one pencil.
c) None of child receives the same number of pencils.
d) If the pencils are given out randomly. What is the probability that there are at least two kids receive the same number of pencils if every kids receives at least one pencil?
I have calculated a) using C(25,20) = 53,130
I calculated b) using C(19,14) = 11,628
for c) I get answer 840 using the following program
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
System.out.println(u+" "+v+" "+x+" "+y+" "+z);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Total Count: "+count);
I see the output:
Count after iteration 1: 144
Count after iteration 2: 144
Count after iteration 3: 120
Count after iteration 4: 120
Count after iteration 5: 96
Count after iteration 6: 72
Count after iteration 7: 48
Count after iteration 8: 48
Count after iteration 9: 24
Count after iteration 10: 24
Total Count: 840
But I don't know how to prove this using discrete mathematics.
Also I need help on d).
probability discrete-mathematics permutations combinations
$endgroup$
add a comment |
$begingroup$
How many ways are there to pass out 20 pencils (assume all the pencils are identical the same) to six children?
Based on the following condition:
a) No restriction. ( i.e. each kid may receive zero to 20 pencils.)
b) Every child receives at least one pencil.
c) None of child receives the same number of pencils.
d) If the pencils are given out randomly. What is the probability that there are at least two kids receive the same number of pencils if every kids receives at least one pencil?
I have calculated a) using C(25,20) = 53,130
I calculated b) using C(19,14) = 11,628
for c) I get answer 840 using the following program
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
System.out.println(u+" "+v+" "+x+" "+y+" "+z);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Total Count: "+count);
I see the output:
Count after iteration 1: 144
Count after iteration 2: 144
Count after iteration 3: 120
Count after iteration 4: 120
Count after iteration 5: 96
Count after iteration 6: 72
Count after iteration 7: 48
Count after iteration 8: 48
Count after iteration 9: 24
Count after iteration 10: 24
Total Count: 840
But I don't know how to prove this using discrete mathematics.
Also I need help on d).
probability discrete-mathematics permutations combinations
$endgroup$
add a comment |
$begingroup$
How many ways are there to pass out 20 pencils (assume all the pencils are identical the same) to six children?
Based on the following condition:
a) No restriction. ( i.e. each kid may receive zero to 20 pencils.)
b) Every child receives at least one pencil.
c) None of child receives the same number of pencils.
d) If the pencils are given out randomly. What is the probability that there are at least two kids receive the same number of pencils if every kids receives at least one pencil?
I have calculated a) using C(25,20) = 53,130
I calculated b) using C(19,14) = 11,628
for c) I get answer 840 using the following program
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
System.out.println(u+" "+v+" "+x+" "+y+" "+z);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Total Count: "+count);
I see the output:
Count after iteration 1: 144
Count after iteration 2: 144
Count after iteration 3: 120
Count after iteration 4: 120
Count after iteration 5: 96
Count after iteration 6: 72
Count after iteration 7: 48
Count after iteration 8: 48
Count after iteration 9: 24
Count after iteration 10: 24
Total Count: 840
But I don't know how to prove this using discrete mathematics.
Also I need help on d).
probability discrete-mathematics permutations combinations
$endgroup$
How many ways are there to pass out 20 pencils (assume all the pencils are identical the same) to six children?
Based on the following condition:
a) No restriction. ( i.e. each kid may receive zero to 20 pencils.)
b) Every child receives at least one pencil.
c) None of child receives the same number of pencils.
d) If the pencils are given out randomly. What is the probability that there are at least two kids receive the same number of pencils if every kids receives at least one pencil?
I have calculated a) using C(25,20) = 53,130
I calculated b) using C(19,14) = 11,628
for c) I get answer 840 using the following program
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
System.out.println(u+" "+v+" "+x+" "+y+" "+z);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Total Count: "+count);
I see the output:
Count after iteration 1: 144
Count after iteration 2: 144
Count after iteration 3: 120
Count after iteration 4: 120
Count after iteration 5: 96
Count after iteration 6: 72
Count after iteration 7: 48
Count after iteration 8: 48
Count after iteration 9: 24
Count after iteration 10: 24
Total Count: 840
But I don't know how to prove this using discrete mathematics.
Also I need help on d).
probability discrete-mathematics permutations combinations
probability discrete-mathematics permutations combinations
edited Mar 19 at 14:17
Maria Mazur
48.9k1260122
48.9k1260122
asked Mar 18 at 21:25
Abhishek PanjabiAbhishek Panjabi
1267
1267
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
a) and b) are correctly calculated.
d) If they all get the different number of pencils, then there would be at least $1+2+3+4+5+6=21$ pencils wich is impossibile.
So the probability that at least two will get the same number of pencil is $1$.
$endgroup$
1
$begingroup$
Every one gets at least one pencil.
$endgroup$
– Maria Mazur
Mar 18 at 21:34
$begingroup$
Have you read what I wrote?
$endgroup$
– Maria Mazur
Mar 18 at 21:45
$begingroup$
can u please elaborate more? and give one or two examples of cases. I can not understand properly.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:48
2
$begingroup$
It is impossibile to everone get differnt number of pencils if each gets at least one pencil. So the probability that at least two will get the same number of pencils is 100%
$endgroup$
– Maria Mazur
Mar 18 at 21:51
$begingroup$
Thank you. That helped.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:52
add a comment |
$begingroup$
I realized that there is no way to generalize the answer for question c). it needs a special solution.
First of all, out of 6 children, 1 child has to get 0 pencils.
After that 5 children are remaining. So, there are just 7 ways to distribute 20 pencils among them, which are:
(1,2,3,4,10)
(1,2,3,5,9)
(1,2,3,6,8)
(1,2,4,5,8)
(1,2,4,6,7)
(1,3,4,5,7)
(2,3,4,5,6)
Now, with these 7 cases, there are 5! ways to arrange them in order.
Therefore, we get 120*7 = 840 ways.
Now, coming back to the first child, to whom we gave 0 pencils. It can be any of the 6 children.
So, 840*6 = 5040 is the final answer.
I could found the first 5 cases by myself, but, as anyone can see that the last two cases are too far and so many cases need to be checked before one can figure those two out. that's why I used a program to compute these. I am also providing that here so that it is helpful for others.
import java.util.*;
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
List<List<Integer>> val = new ArrayList<List<Integer>>();
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
List<Integer> temp = new ArrayList<>();
temp.add(u);
temp.add(v);
temp.add(x);
temp.add(y);
temp.add(z);
Collections.sort(temp);
boolean check = false;
for(List a: val)
Collections.sort(a);
if(a.equals(temp))
check = true;
if(!check)
val.add(temp);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Count: "+count);
System.out.println(val);
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
a) and b) are correctly calculated.
d) If they all get the different number of pencils, then there would be at least $1+2+3+4+5+6=21$ pencils wich is impossibile.
So the probability that at least two will get the same number of pencil is $1$.
$endgroup$
1
$begingroup$
Every one gets at least one pencil.
$endgroup$
– Maria Mazur
Mar 18 at 21:34
$begingroup$
Have you read what I wrote?
$endgroup$
– Maria Mazur
Mar 18 at 21:45
$begingroup$
can u please elaborate more? and give one or two examples of cases. I can not understand properly.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:48
2
$begingroup$
It is impossibile to everone get differnt number of pencils if each gets at least one pencil. So the probability that at least two will get the same number of pencils is 100%
$endgroup$
– Maria Mazur
Mar 18 at 21:51
$begingroup$
Thank you. That helped.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:52
add a comment |
$begingroup$
a) and b) are correctly calculated.
d) If they all get the different number of pencils, then there would be at least $1+2+3+4+5+6=21$ pencils wich is impossibile.
So the probability that at least two will get the same number of pencil is $1$.
$endgroup$
1
$begingroup$
Every one gets at least one pencil.
$endgroup$
– Maria Mazur
Mar 18 at 21:34
$begingroup$
Have you read what I wrote?
$endgroup$
– Maria Mazur
Mar 18 at 21:45
$begingroup$
can u please elaborate more? and give one or two examples of cases. I can not understand properly.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:48
2
$begingroup$
It is impossibile to everone get differnt number of pencils if each gets at least one pencil. So the probability that at least two will get the same number of pencils is 100%
$endgroup$
– Maria Mazur
Mar 18 at 21:51
$begingroup$
Thank you. That helped.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:52
add a comment |
$begingroup$
a) and b) are correctly calculated.
d) If they all get the different number of pencils, then there would be at least $1+2+3+4+5+6=21$ pencils wich is impossibile.
So the probability that at least two will get the same number of pencil is $1$.
$endgroup$
a) and b) are correctly calculated.
d) If they all get the different number of pencils, then there would be at least $1+2+3+4+5+6=21$ pencils wich is impossibile.
So the probability that at least two will get the same number of pencil is $1$.
edited Mar 18 at 21:47
answered Mar 18 at 21:28
Maria MazurMaria Mazur
48.9k1260122
48.9k1260122
1
$begingroup$
Every one gets at least one pencil.
$endgroup$
– Maria Mazur
Mar 18 at 21:34
$begingroup$
Have you read what I wrote?
$endgroup$
– Maria Mazur
Mar 18 at 21:45
$begingroup$
can u please elaborate more? and give one or two examples of cases. I can not understand properly.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:48
2
$begingroup$
It is impossibile to everone get differnt number of pencils if each gets at least one pencil. So the probability that at least two will get the same number of pencils is 100%
$endgroup$
– Maria Mazur
Mar 18 at 21:51
$begingroup$
Thank you. That helped.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:52
add a comment |
1
$begingroup$
Every one gets at least one pencil.
$endgroup$
– Maria Mazur
Mar 18 at 21:34
$begingroup$
Have you read what I wrote?
$endgroup$
– Maria Mazur
Mar 18 at 21:45
$begingroup$
can u please elaborate more? and give one or two examples of cases. I can not understand properly.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:48
2
$begingroup$
It is impossibile to everone get differnt number of pencils if each gets at least one pencil. So the probability that at least two will get the same number of pencils is 100%
$endgroup$
– Maria Mazur
Mar 18 at 21:51
$begingroup$
Thank you. That helped.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:52
1
1
$begingroup$
Every one gets at least one pencil.
$endgroup$
– Maria Mazur
Mar 18 at 21:34
$begingroup$
Every one gets at least one pencil.
$endgroup$
– Maria Mazur
Mar 18 at 21:34
$begingroup$
Have you read what I wrote?
$endgroup$
– Maria Mazur
Mar 18 at 21:45
$begingroup$
Have you read what I wrote?
$endgroup$
– Maria Mazur
Mar 18 at 21:45
$begingroup$
can u please elaborate more? and give one or two examples of cases. I can not understand properly.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:48
$begingroup$
can u please elaborate more? and give one or two examples of cases. I can not understand properly.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:48
2
2
$begingroup$
It is impossibile to everone get differnt number of pencils if each gets at least one pencil. So the probability that at least two will get the same number of pencils is 100%
$endgroup$
– Maria Mazur
Mar 18 at 21:51
$begingroup$
It is impossibile to everone get differnt number of pencils if each gets at least one pencil. So the probability that at least two will get the same number of pencils is 100%
$endgroup$
– Maria Mazur
Mar 18 at 21:51
$begingroup$
Thank you. That helped.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:52
$begingroup$
Thank you. That helped.
$endgroup$
– Abhishek Panjabi
Mar 18 at 21:52
add a comment |
$begingroup$
I realized that there is no way to generalize the answer for question c). it needs a special solution.
First of all, out of 6 children, 1 child has to get 0 pencils.
After that 5 children are remaining. So, there are just 7 ways to distribute 20 pencils among them, which are:
(1,2,3,4,10)
(1,2,3,5,9)
(1,2,3,6,8)
(1,2,4,5,8)
(1,2,4,6,7)
(1,3,4,5,7)
(2,3,4,5,6)
Now, with these 7 cases, there are 5! ways to arrange them in order.
Therefore, we get 120*7 = 840 ways.
Now, coming back to the first child, to whom we gave 0 pencils. It can be any of the 6 children.
So, 840*6 = 5040 is the final answer.
I could found the first 5 cases by myself, but, as anyone can see that the last two cases are too far and so many cases need to be checked before one can figure those two out. that's why I used a program to compute these. I am also providing that here so that it is helpful for others.
import java.util.*;
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
List<List<Integer>> val = new ArrayList<List<Integer>>();
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
List<Integer> temp = new ArrayList<>();
temp.add(u);
temp.add(v);
temp.add(x);
temp.add(y);
temp.add(z);
Collections.sort(temp);
boolean check = false;
for(List a: val)
Collections.sort(a);
if(a.equals(temp))
check = true;
if(!check)
val.add(temp);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Count: "+count);
System.out.println(val);
$endgroup$
add a comment |
$begingroup$
I realized that there is no way to generalize the answer for question c). it needs a special solution.
First of all, out of 6 children, 1 child has to get 0 pencils.
After that 5 children are remaining. So, there are just 7 ways to distribute 20 pencils among them, which are:
(1,2,3,4,10)
(1,2,3,5,9)
(1,2,3,6,8)
(1,2,4,5,8)
(1,2,4,6,7)
(1,3,4,5,7)
(2,3,4,5,6)
Now, with these 7 cases, there are 5! ways to arrange them in order.
Therefore, we get 120*7 = 840 ways.
Now, coming back to the first child, to whom we gave 0 pencils. It can be any of the 6 children.
So, 840*6 = 5040 is the final answer.
I could found the first 5 cases by myself, but, as anyone can see that the last two cases are too far and so many cases need to be checked before one can figure those two out. that's why I used a program to compute these. I am also providing that here so that it is helpful for others.
import java.util.*;
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
List<List<Integer>> val = new ArrayList<List<Integer>>();
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
List<Integer> temp = new ArrayList<>();
temp.add(u);
temp.add(v);
temp.add(x);
temp.add(y);
temp.add(z);
Collections.sort(temp);
boolean check = false;
for(List a: val)
Collections.sort(a);
if(a.equals(temp))
check = true;
if(!check)
val.add(temp);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Count: "+count);
System.out.println(val);
$endgroup$
add a comment |
$begingroup$
I realized that there is no way to generalize the answer for question c). it needs a special solution.
First of all, out of 6 children, 1 child has to get 0 pencils.
After that 5 children are remaining. So, there are just 7 ways to distribute 20 pencils among them, which are:
(1,2,3,4,10)
(1,2,3,5,9)
(1,2,3,6,8)
(1,2,4,5,8)
(1,2,4,6,7)
(1,3,4,5,7)
(2,3,4,5,6)
Now, with these 7 cases, there are 5! ways to arrange them in order.
Therefore, we get 120*7 = 840 ways.
Now, coming back to the first child, to whom we gave 0 pencils. It can be any of the 6 children.
So, 840*6 = 5040 is the final answer.
I could found the first 5 cases by myself, but, as anyone can see that the last two cases are too far and so many cases need to be checked before one can figure those two out. that's why I used a program to compute these. I am also providing that here so that it is helpful for others.
import java.util.*;
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
List<List<Integer>> val = new ArrayList<List<Integer>>();
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
List<Integer> temp = new ArrayList<>();
temp.add(u);
temp.add(v);
temp.add(x);
temp.add(y);
temp.add(z);
Collections.sort(temp);
boolean check = false;
for(List a: val)
Collections.sort(a);
if(a.equals(temp))
check = true;
if(!check)
val.add(temp);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Count: "+count);
System.out.println(val);
$endgroup$
I realized that there is no way to generalize the answer for question c). it needs a special solution.
First of all, out of 6 children, 1 child has to get 0 pencils.
After that 5 children are remaining. So, there are just 7 ways to distribute 20 pencils among them, which are:
(1,2,3,4,10)
(1,2,3,5,9)
(1,2,3,6,8)
(1,2,4,5,8)
(1,2,4,6,7)
(1,3,4,5,7)
(2,3,4,5,6)
Now, with these 7 cases, there are 5! ways to arrange them in order.
Therefore, we get 120*7 = 840 ways.
Now, coming back to the first child, to whom we gave 0 pencils. It can be any of the 6 children.
So, 840*6 = 5040 is the final answer.
I could found the first 5 cases by myself, but, as anyone can see that the last two cases are too far and so many cases need to be checked before one can figure those two out. that's why I used a program to compute these. I am also providing that here so that it is helpful for others.
import java.util.*;
class final34
public static void main(String[] args)
int u,v,x,y,z;
int count = 0;
int last=0;
List<List<Integer>> val = new ArrayList<List<Integer>>();
for(u =1;u<11;u++)
for(v = 1;v<11; v++)
if(u!=v)
for(x = 1;x<11;x++)
if(x!=u && x!=v)
for(y = 1; y<11;y++)
if(y!=u && y!=v && y!=x)
for(z=1;z<11;z++)
if(z!=u && z!=v && z!=x && z!=y)
if(u+v+x+y+z == 20)
List<Integer> temp = new ArrayList<>();
temp.add(u);
temp.add(v);
temp.add(x);
temp.add(y);
temp.add(z);
Collections.sort(temp);
boolean check = false;
for(List a: val)
Collections.sort(a);
if(a.equals(temp))
check = true;
if(!check)
val.add(temp);
count++;
System.out.println("Count after iteration "+u+": "+(count-last));
last = count;
System.out.println("Count: "+count);
System.out.println(val);
answered Mar 18 at 23:54
Abhishek PanjabiAbhishek Panjabi
1267
1267
add a comment |
add a comment |
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