How many ways are there to choose k numbers from first n natural numbers such that any two numbers are at...
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How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?
combinatorics
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
635113
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add a comment |
$begingroup$
How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?
combinatorics
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
635113
$endgroup$
add a comment |
$begingroup$
How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?
combinatorics
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
635113
$endgroup$
How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?
combinatorics
combinatorics
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
635113
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
635113
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
635113
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
635113
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
635113
635113
add a comment |
add a comment |
1 Answer
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$begingroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
answered Jan 7 at 13:10
EuxhenHEuxhenH
472210
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1 Answer
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1 Answer
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$begingroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
answered Jan 7 at 13:10
EuxhenHEuxhenH
472210
$endgroup$
add a comment |
$begingroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
answered Jan 7 at 13:10
EuxhenHEuxhenH
472210
$endgroup$
add a comment |
$begingroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
answered Jan 7 at 13:10
EuxhenHEuxhenH
472210
$endgroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
answered Jan 7 at 13:10
EuxhenHEuxhenH
472210
answered Jan 7 at 13:10
EuxhenHEuxhenH
472210
answered Jan 7 at 13:10
EuxhenHEuxhenH
472210
answered Jan 7 at 13:10
EuxhenHEuxhenH
472210
472210
add a comment |
add a comment |
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