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Split $151$ cakes amongst $3$ people

Combinatorics Split X apples between Y people (with a twist)Formula for Combinations With Replacementunlimited combinationDistributing dimes to 2 groups of people such that each member of one group gets at least onecalculating probability, people in a rowHow many ways can you distribute k different items among n different people if all the items must be shared between r people?Combination with repetition/replacement formulaPlacing 8 people around a table so that 2 people never sit togetherIs there a relation between the triangular numbers and the combinations with repetition?How many ways are there to split $10$ people into two groups of $5$?

2

$begingroup$

$151$ cakes shall be split amongst $3$ people under the condition that no one must have more than $75$. How many combinations are possible?

The solution is $$binom 3+74-13-1$$

which is exactly a combination with replacement.

I don't understand why the number of cakes can be reduced to $74$. There must be some kind of bijectivity between splitting $151$ cakes and the $74$ I don't see.

combinatorics

share|cite|improve this question

edited yesterday

Asaf Karagila♦

306k33437768

asked yesterday

EpsilonDeltaEpsilonDelta

6921615

$endgroup$

add a comment | 

2

$begingroup$

$151$ cakes shall be split amongst $3$ people under the condition that no one must have more than $75$. How many combinations are possible?

The solution is $$binom 3+74-13-1$$

which is exactly a combination with replacement.

I don't understand why the number of cakes can be reduced to $74$. There must be some kind of bijectivity between splitting $151$ cakes and the $74$ I don't see.

combinatorics

share|cite|improve this question

edited yesterday

Asaf Karagila♦

306k33437768

asked yesterday

EpsilonDeltaEpsilonDelta

6921615

$endgroup$

add a comment | 

$begingroup$

$151$ cakes shall be split amongst $3$ people under the condition that no one must have more than $75$. How many combinations are possible?

The solution is $$binom 3+74-13-1$$

which is exactly a combination with replacement.

I don't understand why the number of cakes can be reduced to $74$. There must be some kind of bijectivity between splitting $151$ cakes and the $74$ I don't see.

combinatorics

share|cite|improve this question

edited yesterday

Asaf Karagila♦

306k33437768

asked yesterday

EpsilonDeltaEpsilonDelta

6921615

$endgroup$

share|cite|improve this question

edited yesterday

Asaf Karagila♦

306k33437768

asked yesterday

EpsilonDeltaEpsilonDelta

6921615

share|cite|improve this question

edited yesterday

Asaf Karagila♦

306k33437768

asked yesterday

EpsilonDeltaEpsilonDelta

6921615

edited yesterday

Asaf Karagila♦

306k33437768

edited yesterday

Asaf Karagila♦

306k33437768

asked yesterday

EpsilonDeltaEpsilonDelta

6921615

asked yesterday

EpsilonDeltaEpsilonDelta

6921615

2 Answers
2

active

oldest

votes

5

$begingroup$

The solutions of the system :
$$
x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
$$

is in one-one correspondence with the solutions of the system :
$$
(75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
$$

Which after the change of variable $y_i = 75-x_i$ becomes (since $colorblue75times3 - 151 = 74$):
$$
y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
$$
which by non-negativity has the same solutions as :
$$
y_1+y_2+y_3 = 74 : 0 leq y_i forall i
$$

Now, play the stars-and-bars game to find the number of solutions of this equation.

share|cite|improve this answer

answered yesterday

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

$endgroup$

add a comment | 

2

$begingroup$

"There are no coincidences in mathematics."

The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom752!!!right)=binom75+2-12=binom74+3-13-1$$

This requires some explanation: There is only one way of combining the two numbers to give three numbers less than $75$. For example, if you choose $a$ and $b$, with $age b$, and think of 75 as being split into $a+(75-a)$ and $b+(75-b)$; then combine them to give three acceptable numbers as $a+(75-a+b)+(75-b+1)=151$.

share|cite|improve this answer

edited 18 hours ago

answered yesterday

ChrystomathChrystomath

1,388512

$endgroup$

add a comment | 

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5

$begingroup$

The solutions of the system :
$$
x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
$$

is in one-one correspondence with the solutions of the system :
$$
(75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
$$

Which after the change of variable $y_i = 75-x_i$ becomes (since $colorblue75times3 - 151 = 74$):
$$
y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
$$
which by non-negativity has the same solutions as :
$$
y_1+y_2+y_3 = 74 : 0 leq y_i forall i
$$

Now, play the stars-and-bars game to find the number of solutions of this equation.

share|cite|improve this answer

answered yesterday

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

$endgroup$

add a comment | 

5

$begingroup$

The solutions of the system :
$$
x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
$$

is in one-one correspondence with the solutions of the system :
$$
(75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
$$

Which after the change of variable $y_i = 75-x_i$ becomes (since $colorblue75times3 - 151 = 74$):
$$
y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
$$
which by non-negativity has the same solutions as :
$$
y_1+y_2+y_3 = 74 : 0 leq y_i forall i
$$

Now, play the stars-and-bars game to find the number of solutions of this equation.

share|cite|improve this answer

answered yesterday

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

$endgroup$

add a comment | 

$begingroup$

The solutions of the system :
$$
x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
$$

is in one-one correspondence with the solutions of the system :
$$
(75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
$$

Which after the change of variable $y_i = 75-x_i$ becomes (since $colorblue75times3 - 151 = 74$):
$$
y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
$$
which by non-negativity has the same solutions as :
$$
y_1+y_2+y_3 = 74 : 0 leq y_i forall i
$$

Now, play the stars-and-bars game to find the number of solutions of this equation.

share|cite|improve this answer

answered yesterday

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

$endgroup$

share|cite|improve this answer

answered yesterday

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

answered yesterday

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

answered yesterday

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

2

$begingroup$

"There are no coincidences in mathematics."

The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom752!!!right)=binom75+2-12=binom74+3-13-1$$

This requires some explanation: There is only one way of combining the two numbers to give three numbers less than $75$. For example, if you choose $a$ and $b$, with $age b$, and think of 75 as being split into $a+(75-a)$ and $b+(75-b)$; then combine them to give three acceptable numbers as $a+(75-a+b)+(75-b+1)=151$.

share|cite|improve this answer

edited 18 hours ago

answered yesterday

ChrystomathChrystomath

1,388512

$endgroup$

add a comment | 

2

$begingroup$

"There are no coincidences in mathematics."

The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom752!!!right)=binom75+2-12=binom74+3-13-1$$

This requires some explanation: There is only one way of combining the two numbers to give three numbers less than $75$. For example, if you choose $a$ and $b$, with $age b$, and think of 75 as being split into $a+(75-a)$ and $b+(75-b)$; then combine them to give three acceptable numbers as $a+(75-a+b)+(75-b+1)=151$.

share|cite|improve this answer

edited 18 hours ago

answered yesterday

ChrystomathChrystomath

1,388512

$endgroup$

add a comment | 

$begingroup$

"There are no coincidences in mathematics."

The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom752!!!right)=binom75+2-12=binom74+3-13-1$$

This requires some explanation: There is only one way of combining the two numbers to give three numbers less than $75$. For example, if you choose $a$ and $b$, with $age b$, and think of 75 as being split into $a+(75-a)$ and $b+(75-b)$; then combine them to give three acceptable numbers as $a+(75-a+b)+(75-b+1)=151$.

share|cite|improve this answer

edited 18 hours ago

answered yesterday

ChrystomathChrystomath

1,388512

$endgroup$

share|cite|improve this answer

edited 18 hours ago

answered yesterday

ChrystomathChrystomath

1,388512

answered yesterday

ChrystomathChrystomath

1,388512

answered yesterday

ChrystomathChrystomath

1,388512