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
2,101621
$endgroup$
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
2,101621
$endgroup$
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
2,101621
$endgroup$
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
2,101621
asked Dec 30 '18 at 1:15
MaverickMaverick
2,101621
edited Dec 30 '18 at 1:37
Maverick
asked Dec 30 '18 at 1:15
MaverickMaverick
2,101621
asked Dec 30 '18 at 1:15
MaverickMaverick
2,101621
asked Dec 30 '18 at 1:15
MaverickMaverick
2,101621
2,101621
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056425%2fa-m-of-smallest-elements-of-r-subsets-of-set-1-2-3-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056425%2fa-m-of-smallest-elements-of-r-subsets-of-set-1-2-3-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
