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Combinations and Permutations. Number of integer solutions

1

$begingroup$

Question: How many integer solutions of 

$x_1 + x_2 + x_3 + x_4 = 17$

Satisfy: 

$x_1 ge 0 , x_2 ge 1, x_3 ge 2, x_4 ge 3text ?$

Answer: C(11 + 4 -1, 4-1)

This is the answer the textbook has, can anyone explain how they got that?

Thanks

permutations combinations

share|cite|improve this question

edited Oct 29 '13 at 1:10

Michael Hardy

1

asked Oct 29 '13 at 0:16

asdfasdf

136113

$endgroup$

add a comment | 

1

$begingroup$

Question: How many integer solutions of 

$x_1 + x_2 + x_3 + x_4 = 17$

Satisfy: 

$x_1 ge 0 , x_2 ge 1, x_3 ge 2, x_4 ge 3text ?$

Answer: C(11 + 4 -1, 4-1)

This is the answer the textbook has, can anyone explain how they got that?

Thanks

permutations combinations

share|cite|improve this question

edited Oct 29 '13 at 1:10

Michael Hardy

1

asked Oct 29 '13 at 0:16

asdfasdf

136113

$endgroup$

add a comment | 

$begingroup$

Question: How many integer solutions of 

$x_1 + x_2 + x_3 + x_4 = 17$

Satisfy: 

$x_1 ge 0 , x_2 ge 1, x_3 ge 2, x_4 ge 3text ?$

Answer: C(11 + 4 -1, 4-1)

This is the answer the textbook has, can anyone explain how they got that?

Thanks

permutations combinations

share|cite|improve this question

edited Oct 29 '13 at 1:10

Michael Hardy

1

asked Oct 29 '13 at 0:16

asdfasdf

136113

$endgroup$

Question: How many integer solutions of 

Satisfy: 

share|cite|improve this question

edited Oct 29 '13 at 1:10

Michael Hardy

1

asked Oct 29 '13 at 0:16

asdfasdf

136113

share|cite|improve this question

edited Oct 29 '13 at 1:10

Michael Hardy

1

asked Oct 29 '13 at 0:16

asdfasdf

136113

edited Oct 29 '13 at 1:10

Michael Hardy

1

edited Oct 29 '13 at 1:10

Michael Hardy

1

asked Oct 29 '13 at 0:16

asdfasdf

136113

asked Oct 29 '13 at 0:16

asdfasdf

136113

2 Answers
2

active

oldest

votes

3

$begingroup$

Let $x_2'=x_2-1$, $x_3'=x_3-2$, $x_4'=x_4-3$. We may then rewrite your equation as $$x_1+(x_2'+1)+(x_3'+2)+(x_4'+3)ge 17$$
which we may rewrite again as $$x_1+x_2'+x_3'+x_4'ge 11$$
Only now $x_1, x_2', x_3',x_4'$ are arbitrary nonnegative integers. These are weak compositions of $11$ into $4$ parts, which are given by $$11+4-1choose 4-1$$

share|cite|improve this answer

answered Oct 29 '13 at 0:19

vadim123vadim123

76.6k897191

$endgroup$

add a comment | 

1

$begingroup$

Take $17$ identical marbles.

Temporarily put $6$ of the marbles aside, leaving $11$.

Add $3$ identical dividers to the $11$ marbles.

These $14$ items can be arranged in $14!$ way. Divide by $3!$ and $11!$, because the marbles and dividers are indistinguishable. There are thus $$frac14!3!cdot 11!$$
different arrangements

Now put $1$ of the removed marbles after the first divider, $2$ after the second, and the last $3$ after the third divider. The marble counts represent all the integer solutions to the original equation.

share|cite|improve this answer

answered Oct 29 '13 at 1:23

DJohnMDJohnM

3,2241815

$endgroup$

add a comment | 

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3

$begingroup$

Let $x_2'=x_2-1$, $x_3'=x_3-2$, $x_4'=x_4-3$. We may then rewrite your equation as $$x_1+(x_2'+1)+(x_3'+2)+(x_4'+3)ge 17$$
which we may rewrite again as $$x_1+x_2'+x_3'+x_4'ge 11$$
Only now $x_1, x_2', x_3',x_4'$ are arbitrary nonnegative integers. These are weak compositions of $11$ into $4$ parts, which are given by $$11+4-1choose 4-1$$

share|cite|improve this answer

answered Oct 29 '13 at 0:19

vadim123vadim123

76.6k897191

$endgroup$

add a comment | 

3

$begingroup$

Let $x_2'=x_2-1$, $x_3'=x_3-2$, $x_4'=x_4-3$. We may then rewrite your equation as $$x_1+(x_2'+1)+(x_3'+2)+(x_4'+3)ge 17$$
which we may rewrite again as $$x_1+x_2'+x_3'+x_4'ge 11$$
Only now $x_1, x_2', x_3',x_4'$ are arbitrary nonnegative integers. These are weak compositions of $11$ into $4$ parts, which are given by $$11+4-1choose 4-1$$

share|cite|improve this answer

answered Oct 29 '13 at 0:19

vadim123vadim123

76.6k897191

$endgroup$

add a comment | 

$begingroup$

Let $x_2'=x_2-1$, $x_3'=x_3-2$, $x_4'=x_4-3$. We may then rewrite your equation as $$x_1+(x_2'+1)+(x_3'+2)+(x_4'+3)ge 17$$
which we may rewrite again as $$x_1+x_2'+x_3'+x_4'ge 11$$
Only now $x_1, x_2', x_3',x_4'$ are arbitrary nonnegative integers. These are weak compositions of $11$ into $4$ parts, which are given by $$11+4-1choose 4-1$$

share|cite|improve this answer

answered Oct 29 '13 at 0:19

vadim123vadim123

76.6k897191

$endgroup$

share|cite|improve this answer

answered Oct 29 '13 at 0:19

vadim123vadim123

76.6k897191

answered Oct 29 '13 at 0:19

vadim123vadim123

76.6k897191

answered Oct 29 '13 at 0:19

vadim123vadim123

76.6k897191

1

$begingroup$

Take $17$ identical marbles.

Temporarily put $6$ of the marbles aside, leaving $11$.

Add $3$ identical dividers to the $11$ marbles.

These $14$ items can be arranged in $14!$ way. Divide by $3!$ and $11!$, because the marbles and dividers are indistinguishable. There are thus $$frac14!3!cdot 11!$$
different arrangements

Now put $1$ of the removed marbles after the first divider, $2$ after the second, and the last $3$ after the third divider. The marble counts represent all the integer solutions to the original equation.

share|cite|improve this answer

answered Oct 29 '13 at 1:23

DJohnMDJohnM

3,2241815

$endgroup$

add a comment | 

1

$begingroup$

Take $17$ identical marbles.

Temporarily put $6$ of the marbles aside, leaving $11$.

Add $3$ identical dividers to the $11$ marbles.

These $14$ items can be arranged in $14!$ way. Divide by $3!$ and $11!$, because the marbles and dividers are indistinguishable. There are thus $$frac14!3!cdot 11!$$
different arrangements

Now put $1$ of the removed marbles after the first divider, $2$ after the second, and the last $3$ after the third divider. The marble counts represent all the integer solutions to the original equation.

share|cite|improve this answer

answered Oct 29 '13 at 1:23

DJohnMDJohnM

3,2241815

$endgroup$

add a comment | 

$begingroup$

Take $17$ identical marbles.

Temporarily put $6$ of the marbles aside, leaving $11$.

Add $3$ identical dividers to the $11$ marbles.

These $14$ items can be arranged in $14!$ way. Divide by $3!$ and $11!$, because the marbles and dividers are indistinguishable. There are thus $$frac14!3!cdot 11!$$
different arrangements

Now put $1$ of the removed marbles after the first divider, $2$ after the second, and the last $3$ after the third divider. The marble counts represent all the integer solutions to the original equation.

share|cite|improve this answer

answered Oct 29 '13 at 1:23

DJohnMDJohnM

3,2241815

$endgroup$

share|cite|improve this answer

answered Oct 29 '13 at 1:23

DJohnMDJohnM

3,2241815

answered Oct 29 '13 at 1:23

DJohnMDJohnM

3,2241815

answered Oct 29 '13 at 1:23

DJohnMDJohnM

3,2241815