Consider the identity kC(n,k) = nC(n-1,k-1) where 1 <= k <= n, provy the identity by induction on n, using Pascal's identity Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Uniqueness Proof, Discrete Math HelpEvaluate C(21,2) using Pascal's Identity??riddles and compound propositionInduction proof using Pascal's Identity: $binomn0+binomni+…+binomnn=2^n$Defining a function as a compound propositionWhat is the probability that $2$ of $4$ captured elk have been tagged?Pascal's triangle identityIs this relation an equivalence relation? $xRy :Leftrightarrow x cdot y$ is a square numberIntuitive understanding of Pascal's IdentityConsider the relation $U$ on the set $mathbbZ^*$ is defined as $aUb iff a|b$
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Consider the identity kC(n,k) = nC(n-1,k-1) where 1
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Uniqueness Proof, Discrete Math HelpEvaluate C(21,2) using Pascal's Identity??riddles and compound propositionInduction proof using Pascal's Identity: $binomn0+binomni+…+binomnn=2^n$Defining a function as a compound propositionWhat is the probability that $2$ of $4$ captured elk have been tagged?Pascal's triangle identityIs this relation an equivalence relation? $xRy :Leftrightarrow x cdot y$ is a square numberIntuitive understanding of Pascal's IdentityConsider the relation $U$ on the set $mathbbZ^*$ is defined as $aUb iff a|b$
$begingroup$
I've tried looking everywhere to get a clear understanding of the answer, however I am at a loss. The book says if n=1 then k=1. Assume the identity is true for n-1 we will shot it for n. If k=n, then both sides equal n. Otherwise k ≤ n-1.
kC(n,k) = k(C(n-1,k) + C(n-1,k-1))
=kC(n-1,k) + (k-1)C(n-1,k-1)+C(n-1,k-1)
=(n-1)C(n-2,k-1)+(n-1)C(n-2,k-2)+C(n-1,k-1)
=(n-1)[C(n-2,k-1)+C(n-2,k-2)]+C(n-1,k-1)
(this isn't complete because I get how they get nC(n-1,k-1))
Anyway, I understand the first two rows. However for the last two I don't know where the (n-1), (n-2),(k-2) parts came from.
Sorry if I'm wrong about this part, but is the reason for replacing k with (n-1) because k = n-1?
discrete-mathematics
asked Mar 31 at 23:56
ChrisD93ChrisD93
263
$endgroup$
add a comment |
$begingroup$
I've tried looking everywhere to get a clear understanding of the answer, however I am at a loss. The book says if n=1 then k=1. Assume the identity is true for n-1 we will shot it for n. If k=n, then both sides equal n. Otherwise k ≤ n-1.
kC(n,k) = k(C(n-1,k) + C(n-1,k-1))
=kC(n-1,k) + (k-1)C(n-1,k-1)+C(n-1,k-1)
=(n-1)C(n-2,k-1)+(n-1)C(n-2,k-2)+C(n-1,k-1)
=(n-1)[C(n-2,k-1)+C(n-2,k-2)]+C(n-1,k-1)
(this isn't complete because I get how they get nC(n-1,k-1))
Anyway, I understand the first two rows. However for the last two I don't know where the (n-1), (n-2),(k-2) parts came from.
Sorry if I'm wrong about this part, but is the reason for replacing k with (n-1) because k = n-1?
discrete-mathematics
asked Mar 31 at 23:56
ChrisD93ChrisD93
263
$endgroup$
add a comment |
$begingroup$
I've tried looking everywhere to get a clear understanding of the answer, however I am at a loss. The book says if n=1 then k=1. Assume the identity is true for n-1 we will shot it for n. If k=n, then both sides equal n. Otherwise k ≤ n-1.
kC(n,k) = k(C(n-1,k) + C(n-1,k-1))
=kC(n-1,k) + (k-1)C(n-1,k-1)+C(n-1,k-1)
=(n-1)C(n-2,k-1)+(n-1)C(n-2,k-2)+C(n-1,k-1)
=(n-1)[C(n-2,k-1)+C(n-2,k-2)]+C(n-1,k-1)
(this isn't complete because I get how they get nC(n-1,k-1))
Anyway, I understand the first two rows. However for the last two I don't know where the (n-1), (n-2),(k-2) parts came from.
Sorry if I'm wrong about this part, but is the reason for replacing k with (n-1) because k = n-1?
discrete-mathematics
asked Mar 31 at 23:56
ChrisD93ChrisD93
263
$endgroup$
I've tried looking everywhere to get a clear understanding of the answer, however I am at a loss. The book says if n=1 then k=1. Assume the identity is true for n-1 we will shot it for n. If k=n, then both sides equal n. Otherwise k ≤ n-1.
kC(n,k) = k(C(n-1,k) + C(n-1,k-1))
=kC(n-1,k) + (k-1)C(n-1,k-1)+C(n-1,k-1)
=(n-1)C(n-2,k-1)+(n-1)C(n-2,k-2)+C(n-1,k-1)
=(n-1)[C(n-2,k-1)+C(n-2,k-2)]+C(n-1,k-1)
(this isn't complete because I get how they get nC(n-1,k-1))
Anyway, I understand the first two rows. However for the last two I don't know where the (n-1), (n-2),(k-2) parts came from.
Sorry if I'm wrong about this part, but is the reason for replacing k with (n-1) because k = n-1?
discrete-mathematics
discrete-mathematics
asked Mar 31 at 23:56
ChrisD93ChrisD93
263
asked Mar 31 at 23:56
ChrisD93ChrisD93
263
asked Mar 31 at 23:56
ChrisD93ChrisD93
263
asked Mar 31 at 23:56
ChrisD93ChrisD93
263
asked Mar 31 at 23:56
ChrisD93ChrisD93
263
263
add a comment |
add a comment |
2 Answers
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active
oldest
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$begingroup$
(I know you asked for induction, but I can never resist a combinatorial proof.)
Suppose you have $n$ people and want to form a committee of $k$ people with president. We can do this in two different ways.
One way is to first form a committee without a president, which can be done in $C(n,k)$ ways. Out of the chosen $k$ people, select one of them to be the president. From this perspective, there are $k C(n,k)$ ways to form the committee.
Another way is to first select a president out of all $n$ people. Now that you have a president, choose an additional $k-1$ people to serve on the committee out of the remaining $n-1$ people, which can be done in $C(n-1,k-1)$ ways. From this perspective, there are $nC(n-1,k-1)$ ways to form the committee.
Since the two expressions enumerate the same object, we conclude that $kC(n,k) = nC(n-1,k-1)$.
answered Apr 1 at 0:35
Austin MohrAustin Mohr
20.9k35299
$endgroup$
$begingroup$
I would have went with the combinatorial proof since I find it easier, but I have a feeling my teacher might ask about the induction part .-.
$endgroup$
– ChrisD93
Apr 1 at 0:47
add a comment |
$begingroup$
begineqnarray*
kbinomnk= k binomn-1k-1+k binomn-1k.
endeqnarray*
The second term on the right can be rewritten (using the symmetry of binomial coefficients)
begineqnarray*
k binomn-1k= (n-1) binomn-2k-1 =(n-1) binomn-2n-k-1=(n-k) binomn-1n-k=(n-k) binomn-1k-1.
endeqnarray*
answered Apr 1 at 0:10
Donald SplutterwitDonald Splutterwit
23.1k21446
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
(I know you asked for induction, but I can never resist a combinatorial proof.)
Suppose you have $n$ people and want to form a committee of $k$ people with president. We can do this in two different ways.
One way is to first form a committee without a president, which can be done in $C(n,k)$ ways. Out of the chosen $k$ people, select one of them to be the president. From this perspective, there are $k C(n,k)$ ways to form the committee.
Another way is to first select a president out of all $n$ people. Now that you have a president, choose an additional $k-1$ people to serve on the committee out of the remaining $n-1$ people, which can be done in $C(n-1,k-1)$ ways. From this perspective, there are $nC(n-1,k-1)$ ways to form the committee.
Since the two expressions enumerate the same object, we conclude that $kC(n,k) = nC(n-1,k-1)$.
answered Apr 1 at 0:35
Austin MohrAustin Mohr
20.9k35299
$endgroup$
$begingroup$
I would have went with the combinatorial proof since I find it easier, but I have a feeling my teacher might ask about the induction part .-.
$endgroup$
– ChrisD93
Apr 1 at 0:47
add a comment |
$begingroup$
(I know you asked for induction, but I can never resist a combinatorial proof.)
Suppose you have $n$ people and want to form a committee of $k$ people with president. We can do this in two different ways.
One way is to first form a committee without a president, which can be done in $C(n,k)$ ways. Out of the chosen $k$ people, select one of them to be the president. From this perspective, there are $k C(n,k)$ ways to form the committee.
Another way is to first select a president out of all $n$ people. Now that you have a president, choose an additional $k-1$ people to serve on the committee out of the remaining $n-1$ people, which can be done in $C(n-1,k-1)$ ways. From this perspective, there are $nC(n-1,k-1)$ ways to form the committee.
Since the two expressions enumerate the same object, we conclude that $kC(n,k) = nC(n-1,k-1)$.
answered Apr 1 at 0:35
Austin MohrAustin Mohr
20.9k35299
$endgroup$
$begingroup$
I would have went with the combinatorial proof since I find it easier, but I have a feeling my teacher might ask about the induction part .-.
$endgroup$
– ChrisD93
Apr 1 at 0:47
add a comment |
$begingroup$
(I know you asked for induction, but I can never resist a combinatorial proof.)
Suppose you have $n$ people and want to form a committee of $k$ people with president. We can do this in two different ways.
One way is to first form a committee without a president, which can be done in $C(n,k)$ ways. Out of the chosen $k$ people, select one of them to be the president. From this perspective, there are $k C(n,k)$ ways to form the committee.
Another way is to first select a president out of all $n$ people. Now that you have a president, choose an additional $k-1$ people to serve on the committee out of the remaining $n-1$ people, which can be done in $C(n-1,k-1)$ ways. From this perspective, there are $nC(n-1,k-1)$ ways to form the committee.
Since the two expressions enumerate the same object, we conclude that $kC(n,k) = nC(n-1,k-1)$.
answered Apr 1 at 0:35
Austin MohrAustin Mohr
20.9k35299
$endgroup$
(I know you asked for induction, but I can never resist a combinatorial proof.)
Suppose you have $n$ people and want to form a committee of $k$ people with president. We can do this in two different ways.
One way is to first form a committee without a president, which can be done in $C(n,k)$ ways. Out of the chosen $k$ people, select one of them to be the president. From this perspective, there are $k C(n,k)$ ways to form the committee.
Another way is to first select a president out of all $n$ people. Now that you have a president, choose an additional $k-1$ people to serve on the committee out of the remaining $n-1$ people, which can be done in $C(n-1,k-1)$ ways. From this perspective, there are $nC(n-1,k-1)$ ways to form the committee.
Since the two expressions enumerate the same object, we conclude that $kC(n,k) = nC(n-1,k-1)$.
answered Apr 1 at 0:35
Austin MohrAustin Mohr
20.9k35299
answered Apr 1 at 0:35
Austin MohrAustin Mohr
20.9k35299
answered Apr 1 at 0:35
Austin MohrAustin Mohr
20.9k35299
answered Apr 1 at 0:35
Austin MohrAustin Mohr
20.9k35299
20.9k35299
$begingroup$
I would have went with the combinatorial proof since I find it easier, but I have a feeling my teacher might ask about the induction part .-.
$endgroup$
– ChrisD93
Apr 1 at 0:47
add a comment |
$begingroup$
I would have went with the combinatorial proof since I find it easier, but I have a feeling my teacher might ask about the induction part .-.
$endgroup$
– ChrisD93
Apr 1 at 0:47
$begingroup$
I would have went with the combinatorial proof since I find it easier, but I have a feeling my teacher might ask about the induction part .-.
$endgroup$
– ChrisD93
Apr 1 at 0:47
$begingroup$
I would have went with the combinatorial proof since I find it easier, but I have a feeling my teacher might ask about the induction part .-.
$endgroup$
– ChrisD93
Apr 1 at 0:47
add a comment |
$begingroup$
begineqnarray*
kbinomnk= k binomn-1k-1+k binomn-1k.
endeqnarray*
The second term on the right can be rewritten (using the symmetry of binomial coefficients)
begineqnarray*
k binomn-1k= (n-1) binomn-2k-1 =(n-1) binomn-2n-k-1=(n-k) binomn-1n-k=(n-k) binomn-1k-1.
endeqnarray*
answered Apr 1 at 0:10
Donald SplutterwitDonald Splutterwit
23.1k21446
$endgroup$
add a comment |
$begingroup$
begineqnarray*
kbinomnk= k binomn-1k-1+k binomn-1k.
endeqnarray*
The second term on the right can be rewritten (using the symmetry of binomial coefficients)
begineqnarray*
k binomn-1k= (n-1) binomn-2k-1 =(n-1) binomn-2n-k-1=(n-k) binomn-1n-k=(n-k) binomn-1k-1.
endeqnarray*
answered Apr 1 at 0:10
Donald SplutterwitDonald Splutterwit
23.1k21446
$endgroup$
add a comment |
$begingroup$
begineqnarray*
kbinomnk= k binomn-1k-1+k binomn-1k.
endeqnarray*
The second term on the right can be rewritten (using the symmetry of binomial coefficients)
begineqnarray*
k binomn-1k= (n-1) binomn-2k-1 =(n-1) binomn-2n-k-1=(n-k) binomn-1n-k=(n-k) binomn-1k-1.
endeqnarray*
answered Apr 1 at 0:10
Donald SplutterwitDonald Splutterwit
23.1k21446
$endgroup$
begineqnarray*
kbinomnk= k binomn-1k-1+k binomn-1k.
endeqnarray*
The second term on the right can be rewritten (using the symmetry of binomial coefficients)
begineqnarray*
k binomn-1k= (n-1) binomn-2k-1 =(n-1) binomn-2n-k-1=(n-k) binomn-1n-k=(n-k) binomn-1k-1.
endeqnarray*
answered Apr 1 at 0:10
Donald SplutterwitDonald Splutterwit
23.1k21446
answered Apr 1 at 0:10
Donald SplutterwitDonald Splutterwit
23.1k21446
answered Apr 1 at 0:10
Donald SplutterwitDonald Splutterwit
23.1k21446
answered Apr 1 at 0:10
Donald SplutterwitDonald Splutterwit
23.1k21446
23.1k21446
add a comment |
add a comment |
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