Mixing
Find closed form of sum of fraction of binomial coefficients
$begingroup$
can somebody give me a hint for this exercise, where I have to find the specific closed form?
$sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}, m,ninmathbb{N}$ and $mleq n$
What I have done so far:
$sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}} = sum_{k=0}^m frac{frac{m!}{(m-k)!cdot k!}}{frac{n!}{(n-k)!cdot k!}} = sum_{k=0}^m frac{m!}{n!}cdot frac{(n-k)!}{(m-k)!} = frac{m!}{n!}cdot sum_{k=0}^m frac{(n-k)!}{(m-k)!}$
Look at example 1: $n=8, m=5$
$frac{5!}{8!}cdot(frac{8!}{5!} + frac{7!}{4!} +frac{6!}{3!} +frac{5!}{2!} + frac{4!}{1!} +frac{3!}{0!}) = \frac{5!}{8!} cdot (8cdot7cdot6+7cdot6cdot5+6cdot5cdot4+5cdot4cdot3+4cdot3cdot2+3cdot2cdot1) = \
frac{5!}{8!} cdot (336+210+120+60+24+6) = frac{5!}{8!}cdot 756 = 2.25$
I can't find a pattern.
Edit 1: Maybe there is an approach with recursion.
Edit 2: Okay I found the solution.
$sum_{k=0}^m frac{m!}{n!}cdot sum_{k=0}^m frac{(n-k)!}{(m-k)!}= frac{m!}{n!}cdotfrac{1}{4}cdot tcdot(t+1)cdot(t+2)cdot(t+3)$ with $t=(n-m)!$
Edit 3: This formula works well for example 1, but fails for example 2: $n=9, m=3$
$frac{3!}{9!}cdot(frac{9!}{3!} + frac{8!}{2!} +frac{7!}{1!} +frac{6!}{0!}) = \frac{3!}{9!} cdot (9cdot8cdot7cdot6cdot5cdot4+8cdot7cdot6cdot5cdot4cdot3+7cdot6cdot5cdot4cdot3cdot2+6cdot5cdot4cdot3cdot2cdot1) = \
frac{3!}{9!} cdot (60480+20160+5040+720) = frac{3!}{9!}cdot 86400 = 1.428$
So I have still no general solution. Can someone help?
summation binomial-coefficients closed-form
edited Jan 14 at 15:14
Martin Sleziak
44.8k10118272
asked Jan 13 at 18:38
MatthiasMatthias
253
$endgroup$
add a comment |
$begingroup$
can somebody give me a hint for this exercise, where I have to find the specific closed form?
$sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}, m,ninmathbb{N}$ and $mleq n$
What I have done so far:
$sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}} = sum_{k=0}^m frac{frac{m!}{(m-k)!cdot k!}}{frac{n!}{(n-k)!cdot k!}} = sum_{k=0}^m frac{m!}{n!}cdot frac{(n-k)!}{(m-k)!} = frac{m!}{n!}cdot sum_{k=0}^m frac{(n-k)!}{(m-k)!}$
Look at example 1: $n=8, m=5$
$frac{5!}{8!}cdot(frac{8!}{5!} + frac{7!}{4!} +frac{6!}{3!} +frac{5!}{2!} + frac{4!}{1!} +frac{3!}{0!}) = \frac{5!}{8!} cdot (8cdot7cdot6+7cdot6cdot5+6cdot5cdot4+5cdot4cdot3+4cdot3cdot2+3cdot2cdot1) = \
frac{5!}{8!} cdot (336+210+120+60+24+6) = frac{5!}{8!}cdot 756 = 2.25$
I can't find a pattern.
Edit 1: Maybe there is an approach with recursion.
Edit 2: Okay I found the solution.
$sum_{k=0}^m frac{m!}{n!}cdot sum_{k=0}^m frac{(n-k)!}{(m-k)!}= frac{m!}{n!}cdotfrac{1}{4}cdot tcdot(t+1)cdot(t+2)cdot(t+3)$ with $t=(n-m)!$
Edit 3: This formula works well for example 1, but fails for example 2: $n=9, m=3$
$frac{3!}{9!}cdot(frac{9!}{3!} + frac{8!}{2!} +frac{7!}{1!} +frac{6!}{0!}) = \frac{3!}{9!} cdot (9cdot8cdot7cdot6cdot5cdot4+8cdot7cdot6cdot5cdot4cdot3+7cdot6cdot5cdot4cdot3cdot2+6cdot5cdot4cdot3cdot2cdot1) = \
frac{3!}{9!} cdot (60480+20160+5040+720) = frac{3!}{9!}cdot 86400 = 1.428$
So I have still no general solution. Can someone help?
summation binomial-coefficients closed-form
edited Jan 14 at 15:14
Martin Sleziak
44.8k10118272
asked Jan 13 at 18:38
MatthiasMatthias
253
$endgroup$
$begingroup$
Hope the following combinatorial problem help you: given $n$ marbles $m$ of whom are black and the rest are red, what's the probability of choosing at most $m$ marbles all of which being black?
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:38
add a comment |
$begingroup$
can somebody give me a hint for this exercise, where I have to find the specific closed form?
$sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}, m,ninmathbb{N}$ and $mleq n$
What I have done so far:
$sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}} = sum_{k=0}^m frac{frac{m!}{(m-k)!cdot k!}}{frac{n!}{(n-k)!cdot k!}} = sum_{k=0}^m frac{m!}{n!}cdot frac{(n-k)!}{(m-k)!} = frac{m!}{n!}cdot sum_{k=0}^m frac{(n-k)!}{(m-k)!}$
Look at example 1: $n=8, m=5$
$frac{5!}{8!}cdot(frac{8!}{5!} + frac{7!}{4!} +frac{6!}{3!} +frac{5!}{2!} + frac{4!}{1!} +frac{3!}{0!}) = \frac{5!}{8!} cdot (8cdot7cdot6+7cdot6cdot5+6cdot5cdot4+5cdot4cdot3+4cdot3cdot2+3cdot2cdot1) = \
frac{5!}{8!} cdot (336+210+120+60+24+6) = frac{5!}{8!}cdot 756 = 2.25$
I can't find a pattern.
Edit 1: Maybe there is an approach with recursion.
Edit 2: Okay I found the solution.
$sum_{k=0}^m frac{m!}{n!}cdot sum_{k=0}^m frac{(n-k)!}{(m-k)!}= frac{m!}{n!}cdotfrac{1}{4}cdot tcdot(t+1)cdot(t+2)cdot(t+3)$ with $t=(n-m)!$
Edit 3: This formula works well for example 1, but fails for example 2: $n=9, m=3$
$frac{3!}{9!}cdot(frac{9!}{3!} + frac{8!}{2!} +frac{7!}{1!} +frac{6!}{0!}) = \frac{3!}{9!} cdot (9cdot8cdot7cdot6cdot5cdot4+8cdot7cdot6cdot5cdot4cdot3+7cdot6cdot5cdot4cdot3cdot2+6cdot5cdot4cdot3cdot2cdot1) = \
frac{3!}{9!} cdot (60480+20160+5040+720) = frac{3!}{9!}cdot 86400 = 1.428$
So I have still no general solution. Can someone help?
summation binomial-coefficients closed-form
edited Jan 14 at 15:14
Martin Sleziak
44.8k10118272
asked Jan 13 at 18:38
MatthiasMatthias
253
$endgroup$
can somebody give me a hint for this exercise, where I have to find the specific closed form?
$sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}, m,ninmathbb{N}$ and $mleq n$
What I have done so far:
$sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}} = sum_{k=0}^m frac{frac{m!}{(m-k)!cdot k!}}{frac{n!}{(n-k)!cdot k!}} = sum_{k=0}^m frac{m!}{n!}cdot frac{(n-k)!}{(m-k)!} = frac{m!}{n!}cdot sum_{k=0}^m frac{(n-k)!}{(m-k)!}$
Look at example 1: $n=8, m=5$
$frac{5!}{8!}cdot(frac{8!}{5!} + frac{7!}{4!} +frac{6!}{3!} +frac{5!}{2!} + frac{4!}{1!} +frac{3!}{0!}) = \frac{5!}{8!} cdot (8cdot7cdot6+7cdot6cdot5+6cdot5cdot4+5cdot4cdot3+4cdot3cdot2+3cdot2cdot1) = \
frac{5!}{8!} cdot (336+210+120+60+24+6) = frac{5!}{8!}cdot 756 = 2.25$
I can't find a pattern.
Edit 1: Maybe there is an approach with recursion.
Edit 2: Okay I found the solution.
$sum_{k=0}^m frac{m!}{n!}cdot sum_{k=0}^m frac{(n-k)!}{(m-k)!}= frac{m!}{n!}cdotfrac{1}{4}cdot tcdot(t+1)cdot(t+2)cdot(t+3)$ with $t=(n-m)!$
Edit 3: This formula works well for example 1, but fails for example 2: $n=9, m=3$
$frac{3!}{9!}cdot(frac{9!}{3!} + frac{8!}{2!} +frac{7!}{1!} +frac{6!}{0!}) = \frac{3!}{9!} cdot (9cdot8cdot7cdot6cdot5cdot4+8cdot7cdot6cdot5cdot4cdot3+7cdot6cdot5cdot4cdot3cdot2+6cdot5cdot4cdot3cdot2cdot1) = \
frac{3!}{9!} cdot (60480+20160+5040+720) = frac{3!}{9!}cdot 86400 = 1.428$
So I have still no general solution. Can someone help?
summation binomial-coefficients closed-form
summation binomial-coefficients closed-form
edited Jan 14 at 15:14
Martin Sleziak
44.8k10118272
asked Jan 13 at 18:38
MatthiasMatthias
253
edited Jan 14 at 15:14
Martin Sleziak
44.8k10118272
asked Jan 13 at 18:38
MatthiasMatthias
253
edited Jan 14 at 15:14
Martin Sleziak
44.8k10118272
edited Jan 14 at 15:14
Martin Sleziak
44.8k10118272
edited Jan 14 at 15:14
Martin Sleziak
44.8k10118272
44.8k10118272
asked Jan 13 at 18:38
MatthiasMatthias
253
asked Jan 13 at 18:38
MatthiasMatthias
253
asked Jan 13 at 18:38
MatthiasMatthias
253
253
$begingroup$
Hope the following combinatorial problem help you: given $n$ marbles $m$ of whom are black and the rest are red, what's the probability of choosing at most $m$ marbles all of which being black?
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:38
add a comment |
$begingroup$
Hope the following combinatorial problem help you: given $n$ marbles $m$ of whom are black and the rest are red, what's the probability of choosing at most $m$ marbles all of which being black?
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:38
$begingroup$
Hope the following combinatorial problem help you: given $n$ marbles $m$ of whom are black and the rest are red, what's the probability of choosing at most $m$ marbles all of which being black?
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:38
$begingroup$
Hope the following combinatorial problem help you: given $n$ marbles $m$ of whom are black and the rest are red, what's the probability of choosing at most $m$ marbles all of which being black?
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:38
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let
$$S(m,n):=sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}$$
We are going to prove:
$$S(m,n)=frac{n+1}{n+1-m}.tag{1}$$
Obviously (1) is valid for $m=0$ and arbitray $nge 0$. Further, if (1) is valid for $(m-1,n-1)$ it is valid for $(m,n)$ as well:
$$
S(m,n):=1+sum_{k=1}^m frac{frac mkbinom{m-1}{k-1}}{frac nkbinom{n-1}{k-1}}
=1+frac{m}{n} S(m-1,n-1)\
stackrel{I.H.}{=}1+frac{m}{n}frac{n}{n+1-m}=frac{n+1}{n+1-m}.
$$
Thus by induction (1) is valid for arbitrary $0le m le n$.
answered Jan 13 at 20:43
useruser
4,4101929
$endgroup$
$begingroup$
Hi user, your induction is reasonable, but how to get to this expression $S(m,n)=frac{n+1}{n+1-m}$?
$endgroup$
– Matthias
Jan 13 at 20:50
$begingroup$
Play around with small $n$ and $m$ and try finding a pattern.
$endgroup$
– user
Jan 13 at 20:52
$begingroup$
See also: Prove that $sum_{k=0}^{m}frac{binom{m}{k}}{binom{n}{k}}=frac{n+1}{n+1-m}$. Found using Approach0 - for more tips on searching see: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 14 at 15:16
add a comment |
$begingroup$
With $mle n$ we have the sum
$$sum_{k=0}^m {mchoose k} {nchoose k}^{-1}
= sum_{k=0}^m frac{m!}{(m-k)! k!} frac{k! (n-k)!}{n!}
\ = frac{m!}{n!} sum_{k=0}^m frac{(n-k)!}{(m-k)!}
= {nchoose m}^{-1} sum_{k=0}^m {n-kchoose m-k}
\ = {nchoose m}^{-1} sum_{k=0}^m [z^{m-k}] (1+z)^{n-k}
= {nchoose m}^{-1} [z^m] (1+z)^n sum_{k=0}^m z^k (1+z)^{-k}.$$
We may extend $k$ beyond $m$ due to the $z^k$ term and the coefficient
extractor $[z^m]$ in front, getting
$${nchoose m}^{-1} [z^m] (1+z)^n sum_{kge 0} z^k (1+z)^{-k}
= {nchoose m}^{-1} [z^m] (1+z)^n frac{1}{1-z/(1+z)}
\ = {nchoose m}^{-1} [z^m] (1+z)^{n+1}
= {nchoose m}^{-1} {n+1choose m}
= frac{n+1}{n+1-m}.$$
answered Jan 14 at 14:34
Marko RiedelMarko Riedel
40.3k339108
$endgroup$
add a comment |
$begingroup$
Hint:
Use the beta function
begin{align*}
binom{n}{k}^{-1}=(n+1)int_0^1(1-x)^kx^{n-k},dx
end{align*}
answered Jan 13 at 21:17
Markus ScheuerMarkus Scheuer
61.6k457146
$endgroup$
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%2f3072374%2ffind-closed-form-of-sum-of-fraction-of-binomial-coefficients%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let
$$S(m,n):=sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}$$
We are going to prove:
$$S(m,n)=frac{n+1}{n+1-m}.tag{1}$$
Obviously (1) is valid for $m=0$ and arbitray $nge 0$. Further, if (1) is valid for $(m-1,n-1)$ it is valid for $(m,n)$ as well:
$$
S(m,n):=1+sum_{k=1}^m frac{frac mkbinom{m-1}{k-1}}{frac nkbinom{n-1}{k-1}}
=1+frac{m}{n} S(m-1,n-1)\
stackrel{I.H.}{=}1+frac{m}{n}frac{n}{n+1-m}=frac{n+1}{n+1-m}.
$$
Thus by induction (1) is valid for arbitrary $0le m le n$.
answered Jan 13 at 20:43
useruser
4,4101929
$endgroup$
$begingroup$
Hi user, your induction is reasonable, but how to get to this expression $S(m,n)=frac{n+1}{n+1-m}$?
$endgroup$
– Matthias
Jan 13 at 20:50
$begingroup$
Play around with small $n$ and $m$ and try finding a pattern.
$endgroup$
– user
Jan 13 at 20:52
$begingroup$
See also: Prove that $sum_{k=0}^{m}frac{binom{m}{k}}{binom{n}{k}}=frac{n+1}{n+1-m}$. Found using Approach0 - for more tips on searching see: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 14 at 15:16
add a comment |
$begingroup$
Let
$$S(m,n):=sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}$$
We are going to prove:
$$S(m,n)=frac{n+1}{n+1-m}.tag{1}$$
Obviously (1) is valid for $m=0$ and arbitray $nge 0$. Further, if (1) is valid for $(m-1,n-1)$ it is valid for $(m,n)$ as well:
$$
S(m,n):=1+sum_{k=1}^m frac{frac mkbinom{m-1}{k-1}}{frac nkbinom{n-1}{k-1}}
=1+frac{m}{n} S(m-1,n-1)\
stackrel{I.H.}{=}1+frac{m}{n}frac{n}{n+1-m}=frac{n+1}{n+1-m}.
$$
Thus by induction (1) is valid for arbitrary $0le m le n$.
answered Jan 13 at 20:43
useruser
4,4101929
$endgroup$
$begingroup$
Hi user, your induction is reasonable, but how to get to this expression $S(m,n)=frac{n+1}{n+1-m}$?
$endgroup$
– Matthias
Jan 13 at 20:50
$begingroup$
Play around with small $n$ and $m$ and try finding a pattern.
$endgroup$
– user
Jan 13 at 20:52
$begingroup$
See also: Prove that $sum_{k=0}^{m}frac{binom{m}{k}}{binom{n}{k}}=frac{n+1}{n+1-m}$. Found using Approach0 - for more tips on searching see: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 14 at 15:16
add a comment |
$begingroup$
Let
$$S(m,n):=sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}$$
We are going to prove:
$$S(m,n)=frac{n+1}{n+1-m}.tag{1}$$
Obviously (1) is valid for $m=0$ and arbitray $nge 0$. Further, if (1) is valid for $(m-1,n-1)$ it is valid for $(m,n)$ as well:
$$
S(m,n):=1+sum_{k=1}^m frac{frac mkbinom{m-1}{k-1}}{frac nkbinom{n-1}{k-1}}
=1+frac{m}{n} S(m-1,n-1)\
stackrel{I.H.}{=}1+frac{m}{n}frac{n}{n+1-m}=frac{n+1}{n+1-m}.
$$
Thus by induction (1) is valid for arbitrary $0le m le n$.
answered Jan 13 at 20:43
useruser
4,4101929
$endgroup$
Let
$$S(m,n):=sum_{k=0}^m frac{binom{m}{k}}{binom{n}{k}}$$
We are going to prove:
$$S(m,n)=frac{n+1}{n+1-m}.tag{1}$$
Obviously (1) is valid for $m=0$ and arbitray $nge 0$. Further, if (1) is valid for $(m-1,n-1)$ it is valid for $(m,n)$ as well:
$$
S(m,n):=1+sum_{k=1}^m frac{frac mkbinom{m-1}{k-1}}{frac nkbinom{n-1}{k-1}}
=1+frac{m}{n} S(m-1,n-1)\
stackrel{I.H.}{=}1+frac{m}{n}frac{n}{n+1-m}=frac{n+1}{n+1-m}.
$$
Thus by induction (1) is valid for arbitrary $0le m le n$.
answered Jan 13 at 20:43
useruser
4,4101929
edited Jan 13 at 21:58
answered Jan 13 at 20:43
useruser
4,4101929
answered Jan 13 at 20:43
useruser
4,4101929
answered Jan 13 at 20:43
useruser
4,4101929
4,4101929
$begingroup$
Hi user, your induction is reasonable, but how to get to this expression $S(m,n)=frac{n+1}{n+1-m}$?
$endgroup$
– Matthias
Jan 13 at 20:50
$begingroup$
Play around with small $n$ and $m$ and try finding a pattern.
$endgroup$
– user
Jan 13 at 20:52
$begingroup$
See also: Prove that $sum_{k=0}^{m}frac{binom{m}{k}}{binom{n}{k}}=frac{n+1}{n+1-m}$. Found using Approach0 - for more tips on searching see: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 14 at 15:16
add a comment |
$begingroup$
Hi user, your induction is reasonable, but how to get to this expression $S(m,n)=frac{n+1}{n+1-m}$?
$endgroup$
– Matthias
Jan 13 at 20:50
$begingroup$
Play around with small $n$ and $m$ and try finding a pattern.
$endgroup$
– user
Jan 13 at 20:52
$begingroup$
See also: Prove that $sum_{k=0}^{m}frac{binom{m}{k}}{binom{n}{k}}=frac{n+1}{n+1-m}$. Found using Approach0 - for more tips on searching see: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 14 at 15:16
$begingroup$
Hi user, your induction is reasonable, but how to get to this expression $S(m,n)=frac{n+1}{n+1-m}$?
$endgroup$
– Matthias
Jan 13 at 20:50
$begingroup$
Hi user, your induction is reasonable, but how to get to this expression $S(m,n)=frac{n+1}{n+1-m}$?
$endgroup$
– Matthias
Jan 13 at 20:50
$begingroup$
Play around with small $n$ and $m$ and try finding a pattern.
$endgroup$
– user
Jan 13 at 20:52
$begingroup$
Play around with small $n$ and $m$ and try finding a pattern.
$endgroup$
– user
Jan 13 at 20:52
$begingroup$
See also: Prove that $sum_{k=0}^{m}frac{binom{m}{k}}{binom{n}{k}}=frac{n+1}{n+1-m}$. Found using Approach0 - for more tips on searching see: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 14 at 15:16
$begingroup$
See also: Prove that $sum_{k=0}^{m}frac{binom{m}{k}}{binom{n}{k}}=frac{n+1}{n+1-m}$. Found using Approach0 - for more tips on searching see: How to search on this site?
$endgroup$
– Martin Sleziak
Jan 14 at 15:16
add a comment |
$begingroup$
With $mle n$ we have the sum
$$sum_{k=0}^m {mchoose k} {nchoose k}^{-1}
= sum_{k=0}^m frac{m!}{(m-k)! k!} frac{k! (n-k)!}{n!}
\ = frac{m!}{n!} sum_{k=0}^m frac{(n-k)!}{(m-k)!}
= {nchoose m}^{-1} sum_{k=0}^m {n-kchoose m-k}
\ = {nchoose m}^{-1} sum_{k=0}^m [z^{m-k}] (1+z)^{n-k}
= {nchoose m}^{-1} [z^m] (1+z)^n sum_{k=0}^m z^k (1+z)^{-k}.$$
We may extend $k$ beyond $m$ due to the $z^k$ term and the coefficient
extractor $[z^m]$ in front, getting
$${nchoose m}^{-1} [z^m] (1+z)^n sum_{kge 0} z^k (1+z)^{-k}
= {nchoose m}^{-1} [z^m] (1+z)^n frac{1}{1-z/(1+z)}
\ = {nchoose m}^{-1} [z^m] (1+z)^{n+1}
= {nchoose m}^{-1} {n+1choose m}
= frac{n+1}{n+1-m}.$$
answered Jan 14 at 14:34
Marko RiedelMarko Riedel
40.3k339108
$endgroup$
add a comment |
$begingroup$
With $mle n$ we have the sum
$$sum_{k=0}^m {mchoose k} {nchoose k}^{-1}
= sum_{k=0}^m frac{m!}{(m-k)! k!} frac{k! (n-k)!}{n!}
\ = frac{m!}{n!} sum_{k=0}^m frac{(n-k)!}{(m-k)!}
= {nchoose m}^{-1} sum_{k=0}^m {n-kchoose m-k}
\ = {nchoose m}^{-1} sum_{k=0}^m [z^{m-k}] (1+z)^{n-k}
= {nchoose m}^{-1} [z^m] (1+z)^n sum_{k=0}^m z^k (1+z)^{-k}.$$
We may extend $k$ beyond $m$ due to the $z^k$ term and the coefficient
extractor $[z^m]$ in front, getting
$${nchoose m}^{-1} [z^m] (1+z)^n sum_{kge 0} z^k (1+z)^{-k}
= {nchoose m}^{-1} [z^m] (1+z)^n frac{1}{1-z/(1+z)}
\ = {nchoose m}^{-1} [z^m] (1+z)^{n+1}
= {nchoose m}^{-1} {n+1choose m}
= frac{n+1}{n+1-m}.$$
answered Jan 14 at 14:34
Marko RiedelMarko Riedel
40.3k339108
$endgroup$
add a comment |
$begingroup$
With $mle n$ we have the sum
$$sum_{k=0}^m {mchoose k} {nchoose k}^{-1}
= sum_{k=0}^m frac{m!}{(m-k)! k!} frac{k! (n-k)!}{n!}
\ = frac{m!}{n!} sum_{k=0}^m frac{(n-k)!}{(m-k)!}
= {nchoose m}^{-1} sum_{k=0}^m {n-kchoose m-k}
\ = {nchoose m}^{-1} sum_{k=0}^m [z^{m-k}] (1+z)^{n-k}
= {nchoose m}^{-1} [z^m] (1+z)^n sum_{k=0}^m z^k (1+z)^{-k}.$$
We may extend $k$ beyond $m$ due to the $z^k$ term and the coefficient
extractor $[z^m]$ in front, getting
$${nchoose m}^{-1} [z^m] (1+z)^n sum_{kge 0} z^k (1+z)^{-k}
= {nchoose m}^{-1} [z^m] (1+z)^n frac{1}{1-z/(1+z)}
\ = {nchoose m}^{-1} [z^m] (1+z)^{n+1}
= {nchoose m}^{-1} {n+1choose m}
= frac{n+1}{n+1-m}.$$
answered Jan 14 at 14:34
Marko RiedelMarko Riedel
40.3k339108
$endgroup$
With $mle n$ we have the sum
$$sum_{k=0}^m {mchoose k} {nchoose k}^{-1}
= sum_{k=0}^m frac{m!}{(m-k)! k!} frac{k! (n-k)!}{n!}
\ = frac{m!}{n!} sum_{k=0}^m frac{(n-k)!}{(m-k)!}
= {nchoose m}^{-1} sum_{k=0}^m {n-kchoose m-k}
\ = {nchoose m}^{-1} sum_{k=0}^m [z^{m-k}] (1+z)^{n-k}
= {nchoose m}^{-1} [z^m] (1+z)^n sum_{k=0}^m z^k (1+z)^{-k}.$$
We may extend $k$ beyond $m$ due to the $z^k$ term and the coefficient
extractor $[z^m]$ in front, getting
$${nchoose m}^{-1} [z^m] (1+z)^n sum_{kge 0} z^k (1+z)^{-k}
= {nchoose m}^{-1} [z^m] (1+z)^n frac{1}{1-z/(1+z)}
\ = {nchoose m}^{-1} [z^m] (1+z)^{n+1}
= {nchoose m}^{-1} {n+1choose m}
= frac{n+1}{n+1-m}.$$
answered Jan 14 at 14:34
Marko RiedelMarko Riedel
40.3k339108
answered Jan 14 at 14:34
Marko RiedelMarko Riedel
40.3k339108
answered Jan 14 at 14:34
Marko RiedelMarko Riedel
40.3k339108
answered Jan 14 at 14:34
Marko RiedelMarko Riedel
40.3k339108
40.3k339108
add a comment |
add a comment |
$begingroup$
Hint:
Use the beta function
begin{align*}
binom{n}{k}^{-1}=(n+1)int_0^1(1-x)^kx^{n-k},dx
end{align*}
answered Jan 13 at 21:17
Markus ScheuerMarkus Scheuer
61.6k457146
$endgroup$
add a comment |
$begingroup$
Hint:
Use the beta function
begin{align*}
binom{n}{k}^{-1}=(n+1)int_0^1(1-x)^kx^{n-k},dx
end{align*}
answered Jan 13 at 21:17
Markus ScheuerMarkus Scheuer
61.6k457146
$endgroup$
add a comment |
$begingroup$
Hint:
Use the beta function
begin{align*}
binom{n}{k}^{-1}=(n+1)int_0^1(1-x)^kx^{n-k},dx
end{align*}
answered Jan 13 at 21:17
Markus ScheuerMarkus Scheuer
61.6k457146
$endgroup$
Hint:
Use the beta function
begin{align*}
binom{n}{k}^{-1}=(n+1)int_0^1(1-x)^kx^{n-k},dx
end{align*}
answered Jan 13 at 21:17
Markus ScheuerMarkus Scheuer
61.6k457146
answered Jan 13 at 21:17
Markus ScheuerMarkus Scheuer
61.6k457146
answered Jan 13 at 21:17
Markus ScheuerMarkus Scheuer
61.6k457146
answered Jan 13 at 21:17
Markus ScheuerMarkus Scheuer
61.6k457146
61.6k457146
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%2f3072374%2ffind-closed-form-of-sum-of-fraction-of-binomial-coefficients%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
$begingroup$
Hope the following combinatorial problem help you: given $n$ marbles $m$ of whom are black and the rest are red, what's the probability of choosing at most $m$ marbles all of which being black?
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:38