A.M of smallest elements of $r$ subsets of set 1,2,3,…,n
$begingroup$
All possible subsets containing $r$ elements from the set${1,2,3,...,n}$ are formed where $(1leq rleq n)$. What is the arithmetic mean of the smallest elements of these subsets.
My Attempt
The number of such subsets is clearly $binom{n}{r}$.
The number of subsets with $k$ as its smallest element is $binom{n-k}{r-1}$.
So sum of all the smallest elements$$sum_{k=1}^{n-r+1}kbinom{n-k}{r-1}$$
=Coefficient of $x^{r-1}$in expansion$$sum_{k=1}^{n-r+1}k(1+x)^{n-k}$$
=coefficient of $x^{r-1}$ in expansion $$left{frac{(1+x)^{n+1}-(1+x)^r}{x^2}-frac{(n-r+1)(1+x)^{r-1}}{x}right}=binom{n+1}{r+1}$$
So required AM$$=frac{n+1}{r+1}$$
Is there a combinatorical argument to this.
sequences-and-series combinatorics binomial-coefficients binomial-theorem
asked Dec 30 '18 at 1:15
MaverickMaverick
2,182621
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add a comment |
$begingroup$
All possible subsets containing $r$ elements from the set${1,2,3,...,n}$ are formed where $(1leq rleq n)$. What is the arithmetic mean of the smallest elements of these subsets.
My Attempt
The number of such subsets is clearly $binom{n}{r}$.
The number of subsets with $k$ as its smallest element is $binom{n-k}{r-1}$.
So sum of all the smallest elements$$sum_{k=1}^{n-r+1}kbinom{n-k}{r-1}$$
=Coefficient of $x^{r-1}$in expansion$$sum_{k=1}^{n-r+1}k(1+x)^{n-k}$$
=coefficient of $x^{r-1}$ in expansion $$left{frac{(1+x)^{n+1}-(1+x)^r}{x^2}-frac{(n-r+1)(1+x)^{r-1}}{x}right}=binom{n+1}{r+1}$$
So required AM$$=frac{n+1}{r+1}$$
Is there a combinatorical argument to this.
sequences-and-series combinatorics binomial-coefficients binomial-theorem
asked Dec 30 '18 at 1:15
MaverickMaverick
2,182621
$endgroup$
add a comment |
$begingroup$
All possible subsets containing $r$ elements from the set${1,2,3,...,n}$ are formed where $(1leq rleq n)$. What is the arithmetic mean of the smallest elements of these subsets.
My Attempt
The number of such subsets is clearly $binom{n}{r}$.
The number of subsets with $k$ as its smallest element is $binom{n-k}{r-1}$.
So sum of all the smallest elements$$sum_{k=1}^{n-r+1}kbinom{n-k}{r-1}$$
=Coefficient of $x^{r-1}$in expansion$$sum_{k=1}^{n-r+1}k(1+x)^{n-k}$$
=coefficient of $x^{r-1}$ in expansion $$left{frac{(1+x)^{n+1}-(1+x)^r}{x^2}-frac{(n-r+1)(1+x)^{r-1}}{x}right}=binom{n+1}{r+1}$$
So required AM$$=frac{n+1}{r+1}$$
Is there a combinatorical argument to this.
sequences-and-series combinatorics binomial-coefficients binomial-theorem
asked Dec 30 '18 at 1:15
MaverickMaverick
2,182621
$endgroup$
All possible subsets containing $r$ elements from the set${1,2,3,...,n}$ are formed where $(1leq rleq n)$. What is the arithmetic mean of the smallest elements of these subsets.
My Attempt
The number of such subsets is clearly $binom{n}{r}$.
The number of subsets with $k$ as its smallest element is $binom{n-k}{r-1}$.
So sum of all the smallest elements$$sum_{k=1}^{n-r+1}kbinom{n-k}{r-1}$$
=Coefficient of $x^{r-1}$in expansion$$sum_{k=1}^{n-r+1}k(1+x)^{n-k}$$
=coefficient of $x^{r-1}$ in expansion $$left{frac{(1+x)^{n+1}-(1+x)^r}{x^2}-frac{(n-r+1)(1+x)^{r-1}}{x}right}=binom{n+1}{r+1}$$
So required AM$$=frac{n+1}{r+1}$$
Is there a combinatorical argument to this.
sequences-and-series combinatorics binomial-coefficients binomial-theorem
sequences-and-series combinatorics binomial-coefficients binomial-theorem
asked Dec 30 '18 at 1:15
MaverickMaverick
2,182621
asked Dec 30 '18 at 1:15
MaverickMaverick
2,182621
edited Dec 30 '18 at 1:37
Maverick
asked Dec 30 '18 at 1:15
MaverickMaverick
2,182621
asked Dec 30 '18 at 1:15
MaverickMaverick
2,182621
asked Dec 30 '18 at 1:15
MaverickMaverick
2,182621
2,182621
add a comment |
add a comment |
1 Answer
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$begingroup$
We count the number of $r+1$ element subsets of ${0,1,...,n}$. On one hand, this is $binom{n+1}{r+1}$. On the other hand, for each $1 leq k leq n$, consider the number of $r+1$ element subsets of ${0,1,...,n}$ with second smallest element $k$. This can be counted by first taking a $r$ element subset of ${1,...,n}$ with smallest element $k$, then having $k$ ways to choose the last element from ${0,...,k-1}$. Therefore the number is equal to the sum of the smallest element of $r$ element subsets of ${1,...,n}$ with smallest element $k$. Summing over $k$, the sum of smallest elements of $r$ element subsets of ${1,...,n}$ is the number of $r+1$ element subsets of ${0,1,...,n}$, i.e. $binom{n+1}{r+1}$. We then get the A.M. to be $frac{n+1}{r+1}$ upon dividing by $binom{n}{r}$.
answered Dec 30 '18 at 2:17
user630376user630376
813
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
We count the number of $r+1$ element subsets of ${0,1,...,n}$. On one hand, this is $binom{n+1}{r+1}$. On the other hand, for each $1 leq k leq n$, consider the number of $r+1$ element subsets of ${0,1,...,n}$ with second smallest element $k$. This can be counted by first taking a $r$ element subset of ${1,...,n}$ with smallest element $k$, then having $k$ ways to choose the last element from ${0,...,k-1}$. Therefore the number is equal to the sum of the smallest element of $r$ element subsets of ${1,...,n}$ with smallest element $k$. Summing over $k$, the sum of smallest elements of $r$ element subsets of ${1,...,n}$ is the number of $r+1$ element subsets of ${0,1,...,n}$, i.e. $binom{n+1}{r+1}$. We then get the A.M. to be $frac{n+1}{r+1}$ upon dividing by $binom{n}{r}$.
answered Dec 30 '18 at 2:17
user630376user630376
813
$endgroup$
add a comment |
$begingroup$
We count the number of $r+1$ element subsets of ${0,1,...,n}$. On one hand, this is $binom{n+1}{r+1}$. On the other hand, for each $1 leq k leq n$, consider the number of $r+1$ element subsets of ${0,1,...,n}$ with second smallest element $k$. This can be counted by first taking a $r$ element subset of ${1,...,n}$ with smallest element $k$, then having $k$ ways to choose the last element from ${0,...,k-1}$. Therefore the number is equal to the sum of the smallest element of $r$ element subsets of ${1,...,n}$ with smallest element $k$. Summing over $k$, the sum of smallest elements of $r$ element subsets of ${1,...,n}$ is the number of $r+1$ element subsets of ${0,1,...,n}$, i.e. $binom{n+1}{r+1}$. We then get the A.M. to be $frac{n+1}{r+1}$ upon dividing by $binom{n}{r}$.
answered Dec 30 '18 at 2:17
user630376user630376
813
$endgroup$
add a comment |
$begingroup$
We count the number of $r+1$ element subsets of ${0,1,...,n}$. On one hand, this is $binom{n+1}{r+1}$. On the other hand, for each $1 leq k leq n$, consider the number of $r+1$ element subsets of ${0,1,...,n}$ with second smallest element $k$. This can be counted by first taking a $r$ element subset of ${1,...,n}$ with smallest element $k$, then having $k$ ways to choose the last element from ${0,...,k-1}$. Therefore the number is equal to the sum of the smallest element of $r$ element subsets of ${1,...,n}$ with smallest element $k$. Summing over $k$, the sum of smallest elements of $r$ element subsets of ${1,...,n}$ is the number of $r+1$ element subsets of ${0,1,...,n}$, i.e. $binom{n+1}{r+1}$. We then get the A.M. to be $frac{n+1}{r+1}$ upon dividing by $binom{n}{r}$.
answered Dec 30 '18 at 2:17
user630376user630376
813
$endgroup$
We count the number of $r+1$ element subsets of ${0,1,...,n}$. On one hand, this is $binom{n+1}{r+1}$. On the other hand, for each $1 leq k leq n$, consider the number of $r+1$ element subsets of ${0,1,...,n}$ with second smallest element $k$. This can be counted by first taking a $r$ element subset of ${1,...,n}$ with smallest element $k$, then having $k$ ways to choose the last element from ${0,...,k-1}$. Therefore the number is equal to the sum of the smallest element of $r$ element subsets of ${1,...,n}$ with smallest element $k$. Summing over $k$, the sum of smallest elements of $r$ element subsets of ${1,...,n}$ is the number of $r+1$ element subsets of ${0,1,...,n}$, i.e. $binom{n+1}{r+1}$. We then get the A.M. to be $frac{n+1}{r+1}$ upon dividing by $binom{n}{r}$.
answered Dec 30 '18 at 2:17
user630376user630376
813
answered Dec 30 '18 at 2:17
user630376user630376
813
answered Dec 30 '18 at 2:17
user630376user630376
813
answered Dec 30 '18 at 2:17
user630376user630376
813
813
add a comment |
add a comment |
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