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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?

How many ways to divide 5 different books among 3 children so that each child gets at least one book?How many ways can 8 children facing each other in a circle change seats so that each faces a different child.Distribute $n$ identical presents to $k$ children so that each child receives at most $3$ presentsHow Many Ways to distribute six different books among 13 children if no child gets more than one bookCombinatorics: How many ways can 10 candy bars and 8 lollipops be given to five indistinct children?How many different ways can you distribute $5$ apples and $8$ oranges among six children if every child must receive at least one piece of fruit?In how many ways can six different gifts be given to five different children with each child receiving at least one gift?The number of ways in which 10 identical apples can be distributed to six children so that each child receives at least one appleNumber of ways to distribute $10$ different books to three children such that they get $3$, $3$ and $4$Stars and Bars question: You have 12 different toys and 4 children. If each child is to receive 3 toys then how many different ways can this be done?

2

$begingroup$

I'm trying to learn how to count but I'm having some problem.

The problem: In how many ways can $9$ different books and $17$ identical cakes be divided on $6$ (different) children, so that each child receive at least one book and one cake?

I got: The number of ways to divide $6$ different books on $6$ different children is the same as finding all surjective functions from a set $X$, with $|X|=9$ to set $Y$, where $|6|$. The number is $6!×S(9,6)$.

The number of ways to divide $17$ identical cakes on $6$ different children (here is where I am wrong) should in my mind be, after you give out one cake to each child, an unordered sequence with repetition: $binom11+6-16$. However this is wrong, the right answer here is $binom11+6-15$. Can someone please explain that last step to me?

combinatorics discrete-mathematics

share|cite|improve this question

edited Mar 12 at 15:09

N. F. Taussig

44.7k103358

asked Mar 12 at 10:21

BLCAANBLCAAN

624

$endgroup$

add a comment | 

2

$begingroup$

I'm trying to learn how to count but I'm having some problem.

The problem: In how many ways can $9$ different books and $17$ identical cakes be divided on $6$ (different) children, so that each child receive at least one book and one cake?

I got: The number of ways to divide $6$ different books on $6$ different children is the same as finding all surjective functions from a set $X$, with $|X|=9$ to set $Y$, where $|6|$. The number is $6!×S(9,6)$.

The number of ways to divide $17$ identical cakes on $6$ different children (here is where I am wrong) should in my mind be, after you give out one cake to each child, an unordered sequence with repetition: $binom11+6-16$. However this is wrong, the right answer here is $binom11+6-15$. Can someone please explain that last step to me?

combinatorics discrete-mathematics

share|cite|improve this question

edited Mar 12 at 15:09

N. F. Taussig

44.7k103358

asked Mar 12 at 10:21

BLCAANBLCAAN

624

$endgroup$

add a comment | 

$begingroup$

I'm trying to learn how to count but I'm having some problem.

The problem: In how many ways can $9$ different books and $17$ identical cakes be divided on $6$ (different) children, so that each child receive at least one book and one cake?

I got: The number of ways to divide $6$ different books on $6$ different children is the same as finding all surjective functions from a set $X$, with $|X|=9$ to set $Y$, where $|6|$. The number is $6!×S(9,6)$.

The number of ways to divide $17$ identical cakes on $6$ different children (here is where I am wrong) should in my mind be, after you give out one cake to each child, an unordered sequence with repetition: $binom11+6-16$. However this is wrong, the right answer here is $binom11+6-15$. Can someone please explain that last step to me?

combinatorics discrete-mathematics

share|cite|improve this question

edited Mar 12 at 15:09

N. F. Taussig

44.7k103358

asked Mar 12 at 10:21

BLCAANBLCAAN

624

$endgroup$

share|cite|improve this question

edited Mar 12 at 15:09

N. F. Taussig

44.7k103358

asked Mar 12 at 10:21

BLCAANBLCAAN

624

share|cite|improve this question

edited Mar 12 at 15:09

N. F. Taussig

44.7k103358

asked Mar 12 at 10:21

BLCAANBLCAAN

624

edited Mar 12 at 15:09

N. F. Taussig

44.7k103358

edited Mar 12 at 15:09

N. F. Taussig

44.7k103358

asked Mar 12 at 10:21

BLCAANBLCAAN

624

asked Mar 12 at 10:21

BLCAANBLCAAN

624

1 Answer
1

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oldest

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1

$begingroup$

In the stars-and-bars formulation of the second problem, there are $17$ stars (cakes), but only $5$ bars (children) to place in the $17-1=16$ spaces between adjacent stars, because those $5$ bars define $6$ partitions around them (before, between and after) which are identified with the $6$ children.

Then the number of ways to assign cakes to children is the same as the number of ways to choose, among the $16$ spaces, $5$ spaces to hold the bars, hence $binom11+6-15$ and not $binom11+6-16$.

share|cite|improve this answer

answered Mar 12 at 10:39

Parcly TaxelParcly Taxel

44.5k1376109

$endgroup$

add a comment | 

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1

$begingroup$

In the stars-and-bars formulation of the second problem, there are $17$ stars (cakes), but only $5$ bars (children) to place in the $17-1=16$ spaces between adjacent stars, because those $5$ bars define $6$ partitions around them (before, between and after) which are identified with the $6$ children.

Then the number of ways to assign cakes to children is the same as the number of ways to choose, among the $16$ spaces, $5$ spaces to hold the bars, hence $binom11+6-15$ and not $binom11+6-16$.

share|cite|improve this answer

answered Mar 12 at 10:39

Parcly TaxelParcly Taxel

44.5k1376109

$endgroup$

add a comment | 

1

$begingroup$

In the stars-and-bars formulation of the second problem, there are $17$ stars (cakes), but only $5$ bars (children) to place in the $17-1=16$ spaces between adjacent stars, because those $5$ bars define $6$ partitions around them (before, between and after) which are identified with the $6$ children.

Then the number of ways to assign cakes to children is the same as the number of ways to choose, among the $16$ spaces, $5$ spaces to hold the bars, hence $binom11+6-15$ and not $binom11+6-16$.

share|cite|improve this answer

answered Mar 12 at 10:39

Parcly TaxelParcly Taxel

44.5k1376109

$endgroup$

add a comment | 

$begingroup$

In the stars-and-bars formulation of the second problem, there are $17$ stars (cakes), but only $5$ bars (children) to place in the $17-1=16$ spaces between adjacent stars, because those $5$ bars define $6$ partitions around them (before, between and after) which are identified with the $6$ children.

Then the number of ways to assign cakes to children is the same as the number of ways to choose, among the $16$ spaces, $5$ spaces to hold the bars, hence $binom11+6-15$ and not $binom11+6-16$.

share|cite|improve this answer

answered Mar 12 at 10:39

Parcly TaxelParcly Taxel

44.5k1376109

$endgroup$

share|cite|improve this answer

answered Mar 12 at 10:39

Parcly TaxelParcly Taxel

44.5k1376109

answered Mar 12 at 10:39

Parcly TaxelParcly Taxel

44.5k1376109

answered Mar 12 at 10:39

Parcly TaxelParcly Taxel

44.5k1376109