Mixing
Amount of nondecreasing integer k-tuples between 1 and n
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I've come across two different questions (this one and this one) which ask how many different integer tuples $(x_1, dots, x_k)$ exist such that $1 leq x_1 lt x_2 lt cdots lt x_{k-1} lt x_k leq n$ for some given $n$.
The question I've stumbled with is slightly different: instead of the strict inequality, I have $x_i leq x_{i+1}$ for each i.
My attempt
I tried to convert this problem into an equivalent problem with strict inequality so that I could use the formula from the other questions:
Claim. The amount of integer tuples $(x_1, dots, x_k)$ with $1 leq x_1 leq cdots leq x_k leq n$ is equal to the amount of integer tuples $(y_1, dots, y_k)$ such that $1 leq y_1 lt y_2 lt cdots lt y_{k-1} lt y_k leq n + k - 1$.
Proof of the claim. There is a bijection between the set of all valid tuples in the less-than-or-equal-to problem (with upper bound $n$) and the set of the strict-less-than problem (with upper bound $n + k - 1$), since from the former problem we can replace every $x_i$ with $x_i + i - 1$ and get one instance of the new problem.
Therefore the answer would be ${n + k - 1} choose k$
Is my solution correct?
combinatorics proof-verification combinations
asked Jan 12 at 5:11
Pedro APedro A
2,0211827
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add a comment |
$begingroup$
I've come across two different questions (this one and this one) which ask how many different integer tuples $(x_1, dots, x_k)$ exist such that $1 leq x_1 lt x_2 lt cdots lt x_{k-1} lt x_k leq n$ for some given $n$.
The question I've stumbled with is slightly different: instead of the strict inequality, I have $x_i leq x_{i+1}$ for each i.
My attempt
I tried to convert this problem into an equivalent problem with strict inequality so that I could use the formula from the other questions:
Claim. The amount of integer tuples $(x_1, dots, x_k)$ with $1 leq x_1 leq cdots leq x_k leq n$ is equal to the amount of integer tuples $(y_1, dots, y_k)$ such that $1 leq y_1 lt y_2 lt cdots lt y_{k-1} lt y_k leq n + k - 1$.
Proof of the claim. There is a bijection between the set of all valid tuples in the less-than-or-equal-to problem (with upper bound $n$) and the set of the strict-less-than problem (with upper bound $n + k - 1$), since from the former problem we can replace every $x_i$ with $x_i + i - 1$ and get one instance of the new problem.
Therefore the answer would be ${n + k - 1} choose k$
Is my solution correct?
combinatorics proof-verification combinations
asked Jan 12 at 5:11
Pedro APedro A
2,0211827
$endgroup$
1
$begingroup$
That's correct!
$endgroup$
– Mike Earnest
Jan 12 at 6:39
add a comment |
$begingroup$
I've come across two different questions (this one and this one) which ask how many different integer tuples $(x_1, dots, x_k)$ exist such that $1 leq x_1 lt x_2 lt cdots lt x_{k-1} lt x_k leq n$ for some given $n$.
The question I've stumbled with is slightly different: instead of the strict inequality, I have $x_i leq x_{i+1}$ for each i.
My attempt
I tried to convert this problem into an equivalent problem with strict inequality so that I could use the formula from the other questions:
Claim. The amount of integer tuples $(x_1, dots, x_k)$ with $1 leq x_1 leq cdots leq x_k leq n$ is equal to the amount of integer tuples $(y_1, dots, y_k)$ such that $1 leq y_1 lt y_2 lt cdots lt y_{k-1} lt y_k leq n + k - 1$.
Proof of the claim. There is a bijection between the set of all valid tuples in the less-than-or-equal-to problem (with upper bound $n$) and the set of the strict-less-than problem (with upper bound $n + k - 1$), since from the former problem we can replace every $x_i$ with $x_i + i - 1$ and get one instance of the new problem.
Therefore the answer would be ${n + k - 1} choose k$
Is my solution correct?
combinatorics proof-verification combinations
asked Jan 12 at 5:11
Pedro APedro A
2,0211827
$endgroup$
I've come across two different questions (this one and this one) which ask how many different integer tuples $(x_1, dots, x_k)$ exist such that $1 leq x_1 lt x_2 lt cdots lt x_{k-1} lt x_k leq n$ for some given $n$.
The question I've stumbled with is slightly different: instead of the strict inequality, I have $x_i leq x_{i+1}$ for each i.
My attempt
I tried to convert this problem into an equivalent problem with strict inequality so that I could use the formula from the other questions:
Claim. The amount of integer tuples $(x_1, dots, x_k)$ with $1 leq x_1 leq cdots leq x_k leq n$ is equal to the amount of integer tuples $(y_1, dots, y_k)$ such that $1 leq y_1 lt y_2 lt cdots lt y_{k-1} lt y_k leq n + k - 1$.
Proof of the claim. There is a bijection between the set of all valid tuples in the less-than-or-equal-to problem (with upper bound $n$) and the set of the strict-less-than problem (with upper bound $n + k - 1$), since from the former problem we can replace every $x_i$ with $x_i + i - 1$ and get one instance of the new problem.
Therefore the answer would be ${n + k - 1} choose k$
Is my solution correct?
combinatorics proof-verification combinations
combinatorics proof-verification combinations
asked Jan 12 at 5:11
Pedro APedro A
2,0211827
asked Jan 12 at 5:11
Pedro APedro A
2,0211827
asked Jan 12 at 5:11
Pedro APedro A
2,0211827
asked Jan 12 at 5:11
Pedro APedro A
2,0211827
asked Jan 12 at 5:11
Pedro APedro A
2,0211827
2,0211827
1
$begingroup$
That's correct!
$endgroup$
– Mike Earnest
Jan 12 at 6:39
add a comment |
1
$begingroup$
That's correct!
$endgroup$
– Mike Earnest
Jan 12 at 6:39
1
1
$begingroup$
That's correct!
$endgroup$
– Mike Earnest
Jan 12 at 6:39
$begingroup$
That's correct!
$endgroup$
– Mike Earnest
Jan 12 at 6:39
add a comment |
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That's correct!
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– Mike Earnest
Jan 12 at 6:39