how many ways can we choose a committee of 6 animals if there must be at least 3 pigs?
Can anyone help me this problem?
In a group of $2$ cats, $3$ dogs, and $10$ pigs, in how many ways can we choose a committee of $6$ animals if there must be at least $3$ pigs? Assume each kind of animal is identical.
combinatorics combinations generating-functions
asked Mar 2 '17 at 4:02
learning
9151418
add a comment |
Can anyone help me this problem?
In a group of $2$ cats, $3$ dogs, and $10$ pigs, in how many ways can we choose a committee of $6$ animals if there must be at least $3$ pigs? Assume each kind of animal is identical.
combinatorics combinations generating-functions
asked Mar 2 '17 at 4:02
learning
9151418
Are each animal of a kind considered identical or do they all have different names and are considered distinct? If considered identical, consider using stars and bars coupled with inclusion-exclusion. If considered distinct, recommend just approaching directly by cases.
– JMoravitz
Mar 2 '17 at 4:06
add a comment |
Can anyone help me this problem?
In a group of $2$ cats, $3$ dogs, and $10$ pigs, in how many ways can we choose a committee of $6$ animals if there must be at least $3$ pigs? Assume each kind of animal is identical.
combinatorics combinations generating-functions
asked Mar 2 '17 at 4:02
learning
9151418
Can anyone help me this problem?
In a group of $2$ cats, $3$ dogs, and $10$ pigs, in how many ways can we choose a committee of $6$ animals if there must be at least $3$ pigs? Assume each kind of animal is identical.
combinatorics combinations generating-functions
combinatorics combinations generating-functions
asked Mar 2 '17 at 4:02
learning
9151418
asked Mar 2 '17 at 4:02
learning
9151418
edited Mar 2 '17 at 4:25
asked Mar 2 '17 at 4:02
learning
9151418
asked Mar 2 '17 at 4:02
learning
9151418
asked Mar 2 '17 at 4:02
learning
9151418
9151418
Are each animal of a kind considered identical or do they all have different names and are considered distinct? If considered identical, consider using stars and bars coupled with inclusion-exclusion. If considered distinct, recommend just approaching directly by cases.
– JMoravitz
Mar 2 '17 at 4:06
add a comment |
Are each animal of a kind considered identical or do they all have different names and are considered distinct? If considered identical, consider using stars and bars coupled with inclusion-exclusion. If considered distinct, recommend just approaching directly by cases.
– JMoravitz
Mar 2 '17 at 4:06
Are each animal of a kind considered identical or do they all have different names and are considered distinct? If considered identical, consider using stars and bars coupled with inclusion-exclusion. If considered distinct, recommend just approaching directly by cases.
– JMoravitz
Mar 2 '17 at 4:06
Are each animal of a kind considered identical or do they all have different names and are considered distinct? If considered identical, consider using stars and bars coupled with inclusion-exclusion. If considered distinct, recommend just approaching directly by cases.
– JMoravitz
Mar 2 '17 at 4:06
add a comment |
2 Answers
2
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oldest
votes
This is equivalent to the number of integral solutions to the system
$$begin{cases} x_1+x_2+x_3=6\0leq x_1leq 2\ 0leq x_2leq 3\ 3leq x_3color{grey}{leq 10}end{cases}$$
The upper limit on $x_3$ is irrelevant since it is impossible for us to select more pigs than we have in our group of six.
Let $S$ be the set of solutions where we don't care about any of the upper bounds. Let $A_1$ be the set of solutions where (at the very least) the first upperbound condition is violated (and possibly the second one too). Let $A_2$ be the set of solutions where the second upperbound condition is violated.
We wish to count $|Ssetminus (A_1cup A_2)| = |S|-|A_1|-|A_2|+|A_1cap A_2|$
What does it mean for an upper bound condition? Well, $A_1$ is when the condition $x_1leq 2$ is false, i.e. $x_1>2$ i.e. $3leq x_1$. That is to say $|A_1|$ is the number of solutions to
$$begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$$
Remember to continue calculations that the number of solutions to the system
$$begin{cases}x_1+x_2+dots+x_r=n\ 0leq x_i~~forall iend{cases}$$
is seen to be $binom{n+r-1}{r-1}$ via stars and bars.
answered Mar 2 '17 at 5:26
JMoravitz
46.3k33784
But how do we calculate $|A_1|, |A_2| $ and $|A_1|cap|A_2|$
– Osheen Sachdev
Mar 2 '17 at 5:48
1
@OsheenSachdev $|A_1|$ and $|A_2|$ are numbers and cannot be intersected. You mean $|A_1cap A_2|$. Put the system into the form I give at the bottom where the lower bound is zero by making a change of variables. E.g. For $A_1$ it was $begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$. Setting $x_1-3=y_1$, $x_2=y_2$ and $x_3-3=y_3$ we have this system is equivalent to $begin{cases}y_1+y_2+y_3=0\ 0leq y_1\ 0leq y_2\ 0leq y_3end{cases}$, now apply the given formula.
– JMoravitz
Mar 2 '17 at 5:51
Okay got it...thank you so much!
– Osheen Sachdev
Mar 2 '17 at 5:52
add a comment |
Three seats on the committee are for pigs only, and the other three are at-large. Let's count the ways to fill the at-large seats, since there's exactly one way to fill the reserved seats. Each way corresponds to a monomial of degree $3$ in this product:
$$ (1 + c + c^2)(1 + d + d^2 + d^3)(1 + p + p^2 + p^3) enspace, $$
where $c^id^jp^k$ represents a committee with $i$ cats, $j$ dogs, and $k$ pigs. There are nine such terms with non-negative exponents and $i leq 2$:
$$ p^3, dp^2, d^2p, d^3, cp^2, cdp, cd^2, c^2p, c^2d enspace. $$
Hence there are nine ways to form the committee.
answered Mar 2 '17 at 6:19
Fabio Somenzi
6,41221221
add a comment |
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This is equivalent to the number of integral solutions to the system
$$begin{cases} x_1+x_2+x_3=6\0leq x_1leq 2\ 0leq x_2leq 3\ 3leq x_3color{grey}{leq 10}end{cases}$$
The upper limit on $x_3$ is irrelevant since it is impossible for us to select more pigs than we have in our group of six.
Let $S$ be the set of solutions where we don't care about any of the upper bounds. Let $A_1$ be the set of solutions where (at the very least) the first upperbound condition is violated (and possibly the second one too). Let $A_2$ be the set of solutions where the second upperbound condition is violated.
We wish to count $|Ssetminus (A_1cup A_2)| = |S|-|A_1|-|A_2|+|A_1cap A_2|$
What does it mean for an upper bound condition? Well, $A_1$ is when the condition $x_1leq 2$ is false, i.e. $x_1>2$ i.e. $3leq x_1$. That is to say $|A_1|$ is the number of solutions to
$$begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$$
Remember to continue calculations that the number of solutions to the system
$$begin{cases}x_1+x_2+dots+x_r=n\ 0leq x_i~~forall iend{cases}$$
is seen to be $binom{n+r-1}{r-1}$ via stars and bars.
answered Mar 2 '17 at 5:26
JMoravitz
46.3k33784
But how do we calculate $|A_1|, |A_2| $ and $|A_1|cap|A_2|$
– Osheen Sachdev
Mar 2 '17 at 5:48
1
@OsheenSachdev $|A_1|$ and $|A_2|$ are numbers and cannot be intersected. You mean $|A_1cap A_2|$. Put the system into the form I give at the bottom where the lower bound is zero by making a change of variables. E.g. For $A_1$ it was $begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$. Setting $x_1-3=y_1$, $x_2=y_2$ and $x_3-3=y_3$ we have this system is equivalent to $begin{cases}y_1+y_2+y_3=0\ 0leq y_1\ 0leq y_2\ 0leq y_3end{cases}$, now apply the given formula.
– JMoravitz
Mar 2 '17 at 5:51
Okay got it...thank you so much!
– Osheen Sachdev
Mar 2 '17 at 5:52
add a comment |
This is equivalent to the number of integral solutions to the system
$$begin{cases} x_1+x_2+x_3=6\0leq x_1leq 2\ 0leq x_2leq 3\ 3leq x_3color{grey}{leq 10}end{cases}$$
The upper limit on $x_3$ is irrelevant since it is impossible for us to select more pigs than we have in our group of six.
Let $S$ be the set of solutions where we don't care about any of the upper bounds. Let $A_1$ be the set of solutions where (at the very least) the first upperbound condition is violated (and possibly the second one too). Let $A_2$ be the set of solutions where the second upperbound condition is violated.
We wish to count $|Ssetminus (A_1cup A_2)| = |S|-|A_1|-|A_2|+|A_1cap A_2|$
What does it mean for an upper bound condition? Well, $A_1$ is when the condition $x_1leq 2$ is false, i.e. $x_1>2$ i.e. $3leq x_1$. That is to say $|A_1|$ is the number of solutions to
$$begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$$
Remember to continue calculations that the number of solutions to the system
$$begin{cases}x_1+x_2+dots+x_r=n\ 0leq x_i~~forall iend{cases}$$
is seen to be $binom{n+r-1}{r-1}$ via stars and bars.
answered Mar 2 '17 at 5:26
JMoravitz
46.3k33784
But how do we calculate $|A_1|, |A_2| $ and $|A_1|cap|A_2|$
– Osheen Sachdev
Mar 2 '17 at 5:48
1
@OsheenSachdev $|A_1|$ and $|A_2|$ are numbers and cannot be intersected. You mean $|A_1cap A_2|$. Put the system into the form I give at the bottom where the lower bound is zero by making a change of variables. E.g. For $A_1$ it was $begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$. Setting $x_1-3=y_1$, $x_2=y_2$ and $x_3-3=y_3$ we have this system is equivalent to $begin{cases}y_1+y_2+y_3=0\ 0leq y_1\ 0leq y_2\ 0leq y_3end{cases}$, now apply the given formula.
– JMoravitz
Mar 2 '17 at 5:51
Okay got it...thank you so much!
– Osheen Sachdev
Mar 2 '17 at 5:52
add a comment |
This is equivalent to the number of integral solutions to the system
$$begin{cases} x_1+x_2+x_3=6\0leq x_1leq 2\ 0leq x_2leq 3\ 3leq x_3color{grey}{leq 10}end{cases}$$
The upper limit on $x_3$ is irrelevant since it is impossible for us to select more pigs than we have in our group of six.
Let $S$ be the set of solutions where we don't care about any of the upper bounds. Let $A_1$ be the set of solutions where (at the very least) the first upperbound condition is violated (and possibly the second one too). Let $A_2$ be the set of solutions where the second upperbound condition is violated.
We wish to count $|Ssetminus (A_1cup A_2)| = |S|-|A_1|-|A_2|+|A_1cap A_2|$
What does it mean for an upper bound condition? Well, $A_1$ is when the condition $x_1leq 2$ is false, i.e. $x_1>2$ i.e. $3leq x_1$. That is to say $|A_1|$ is the number of solutions to
$$begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$$
Remember to continue calculations that the number of solutions to the system
$$begin{cases}x_1+x_2+dots+x_r=n\ 0leq x_i~~forall iend{cases}$$
is seen to be $binom{n+r-1}{r-1}$ via stars and bars.
answered Mar 2 '17 at 5:26
JMoravitz
46.3k33784
This is equivalent to the number of integral solutions to the system
$$begin{cases} x_1+x_2+x_3=6\0leq x_1leq 2\ 0leq x_2leq 3\ 3leq x_3color{grey}{leq 10}end{cases}$$
The upper limit on $x_3$ is irrelevant since it is impossible for us to select more pigs than we have in our group of six.
Let $S$ be the set of solutions where we don't care about any of the upper bounds. Let $A_1$ be the set of solutions where (at the very least) the first upperbound condition is violated (and possibly the second one too). Let $A_2$ be the set of solutions where the second upperbound condition is violated.
We wish to count $|Ssetminus (A_1cup A_2)| = |S|-|A_1|-|A_2|+|A_1cap A_2|$
What does it mean for an upper bound condition? Well, $A_1$ is when the condition $x_1leq 2$ is false, i.e. $x_1>2$ i.e. $3leq x_1$. That is to say $|A_1|$ is the number of solutions to
$$begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$$
Remember to continue calculations that the number of solutions to the system
$$begin{cases}x_1+x_2+dots+x_r=n\ 0leq x_i~~forall iend{cases}$$
is seen to be $binom{n+r-1}{r-1}$ via stars and bars.
answered Mar 2 '17 at 5:26
JMoravitz
46.3k33784
answered Mar 2 '17 at 5:26
JMoravitz
46.3k33784
answered Mar 2 '17 at 5:26
JMoravitz
46.3k33784
answered Mar 2 '17 at 5:26
JMoravitz
46.3k33784
46.3k33784
But how do we calculate $|A_1|, |A_2| $ and $|A_1|cap|A_2|$
– Osheen Sachdev
Mar 2 '17 at 5:48
1
@OsheenSachdev $|A_1|$ and $|A_2|$ are numbers and cannot be intersected. You mean $|A_1cap A_2|$. Put the system into the form I give at the bottom where the lower bound is zero by making a change of variables. E.g. For $A_1$ it was $begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$. Setting $x_1-3=y_1$, $x_2=y_2$ and $x_3-3=y_3$ we have this system is equivalent to $begin{cases}y_1+y_2+y_3=0\ 0leq y_1\ 0leq y_2\ 0leq y_3end{cases}$, now apply the given formula.
– JMoravitz
Mar 2 '17 at 5:51
Okay got it...thank you so much!
– Osheen Sachdev
Mar 2 '17 at 5:52
add a comment |
But how do we calculate $|A_1|, |A_2| $ and $|A_1|cap|A_2|$
– Osheen Sachdev
Mar 2 '17 at 5:48
1
@OsheenSachdev $|A_1|$ and $|A_2|$ are numbers and cannot be intersected. You mean $|A_1cap A_2|$. Put the system into the form I give at the bottom where the lower bound is zero by making a change of variables. E.g. For $A_1$ it was $begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$. Setting $x_1-3=y_1$, $x_2=y_2$ and $x_3-3=y_3$ we have this system is equivalent to $begin{cases}y_1+y_2+y_3=0\ 0leq y_1\ 0leq y_2\ 0leq y_3end{cases}$, now apply the given formula.
– JMoravitz
Mar 2 '17 at 5:51
Okay got it...thank you so much!
– Osheen Sachdev
Mar 2 '17 at 5:52
But how do we calculate $|A_1|, |A_2| $ and $|A_1|cap|A_2|$
– Osheen Sachdev
Mar 2 '17 at 5:48
But how do we calculate $|A_1|, |A_2| $ and $|A_1|cap|A_2|$
– Osheen Sachdev
Mar 2 '17 at 5:48
1
1
@OsheenSachdev $|A_1|$ and $|A_2|$ are numbers and cannot be intersected. You mean $|A_1cap A_2|$. Put the system into the form I give at the bottom where the lower bound is zero by making a change of variables. E.g. For $A_1$ it was $begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$. Setting $x_1-3=y_1$, $x_2=y_2$ and $x_3-3=y_3$ we have this system is equivalent to $begin{cases}y_1+y_2+y_3=0\ 0leq y_1\ 0leq y_2\ 0leq y_3end{cases}$, now apply the given formula.
– JMoravitz
Mar 2 '17 at 5:51
@OsheenSachdev $|A_1|$ and $|A_2|$ are numbers and cannot be intersected. You mean $|A_1cap A_2|$. Put the system into the form I give at the bottom where the lower bound is zero by making a change of variables. E.g. For $A_1$ it was $begin{cases}x_1+x_2+x_3=6\ 3leq x_1\ 0leq x_2\ 3leq x_3end{cases}$. Setting $x_1-3=y_1$, $x_2=y_2$ and $x_3-3=y_3$ we have this system is equivalent to $begin{cases}y_1+y_2+y_3=0\ 0leq y_1\ 0leq y_2\ 0leq y_3end{cases}$, now apply the given formula.
– JMoravitz
Mar 2 '17 at 5:51
Okay got it...thank you so much!
– Osheen Sachdev
Mar 2 '17 at 5:52
Okay got it...thank you so much!
– Osheen Sachdev
Mar 2 '17 at 5:52
add a comment |
Three seats on the committee are for pigs only, and the other three are at-large. Let's count the ways to fill the at-large seats, since there's exactly one way to fill the reserved seats. Each way corresponds to a monomial of degree $3$ in this product:
$$ (1 + c + c^2)(1 + d + d^2 + d^3)(1 + p + p^2 + p^3) enspace, $$
where $c^id^jp^k$ represents a committee with $i$ cats, $j$ dogs, and $k$ pigs. There are nine such terms with non-negative exponents and $i leq 2$:
$$ p^3, dp^2, d^2p, d^3, cp^2, cdp, cd^2, c^2p, c^2d enspace. $$
Hence there are nine ways to form the committee.
answered Mar 2 '17 at 6:19
Fabio Somenzi
6,41221221
add a comment |
Three seats on the committee are for pigs only, and the other three are at-large. Let's count the ways to fill the at-large seats, since there's exactly one way to fill the reserved seats. Each way corresponds to a monomial of degree $3$ in this product:
$$ (1 + c + c^2)(1 + d + d^2 + d^3)(1 + p + p^2 + p^3) enspace, $$
where $c^id^jp^k$ represents a committee with $i$ cats, $j$ dogs, and $k$ pigs. There are nine such terms with non-negative exponents and $i leq 2$:
$$ p^3, dp^2, d^2p, d^3, cp^2, cdp, cd^2, c^2p, c^2d enspace. $$
Hence there are nine ways to form the committee.
answered Mar 2 '17 at 6:19
Fabio Somenzi
6,41221221
add a comment |
Three seats on the committee are for pigs only, and the other three are at-large. Let's count the ways to fill the at-large seats, since there's exactly one way to fill the reserved seats. Each way corresponds to a monomial of degree $3$ in this product:
$$ (1 + c + c^2)(1 + d + d^2 + d^3)(1 + p + p^2 + p^3) enspace, $$
where $c^id^jp^k$ represents a committee with $i$ cats, $j$ dogs, and $k$ pigs. There are nine such terms with non-negative exponents and $i leq 2$:
$$ p^3, dp^2, d^2p, d^3, cp^2, cdp, cd^2, c^2p, c^2d enspace. $$
Hence there are nine ways to form the committee.
answered Mar 2 '17 at 6:19
Fabio Somenzi
6,41221221
Three seats on the committee are for pigs only, and the other three are at-large. Let's count the ways to fill the at-large seats, since there's exactly one way to fill the reserved seats. Each way corresponds to a monomial of degree $3$ in this product:
$$ (1 + c + c^2)(1 + d + d^2 + d^3)(1 + p + p^2 + p^3) enspace, $$
where $c^id^jp^k$ represents a committee with $i$ cats, $j$ dogs, and $k$ pigs. There are nine such terms with non-negative exponents and $i leq 2$:
$$ p^3, dp^2, d^2p, d^3, cp^2, cdp, cd^2, c^2p, c^2d enspace. $$
Hence there are nine ways to form the committee.
answered Mar 2 '17 at 6:19
Fabio Somenzi
6,41221221
answered Mar 2 '17 at 6:19
Fabio Somenzi
6,41221221
answered Mar 2 '17 at 6:19
Fabio Somenzi
6,41221221
answered Mar 2 '17 at 6:19
Fabio Somenzi
6,41221221
6,41221221
add a comment |
add a comment |
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Are each animal of a kind considered identical or do they all have different names and are considered distinct? If considered identical, consider using stars and bars coupled with inclusion-exclusion. If considered distinct, recommend just approaching directly by cases.
– JMoravitz
Mar 2 '17 at 4:06