Mixing
A problem on Catalan numbers
Suppose two candidates A and B poll the same number of votes, $n$ each,
in an election. The counting of votes is usually done in some arbitrary order and
therefore, during the counting process A may lead for some period and B may lead for
some period. Prove that the number of ways of counting the votes such that A never trails B is
the $n^{th}$ catalan number.
I know that in such questions we are supposed to analyse the problem by dropping a few terms. For example, in this one, we can drop the $2n^{th}$ vote and check it for $2n-1$ votes, but it doesn't seem to be correct since I won't get catalan number. Can someone please help in getting a combinatorial argument/proof for this? Thanks!
combinatorics recurrence-relations catalan-numbers
asked Nov 21 '18 at 6:04
Ankit Kumar
1
add a comment |
Suppose two candidates A and B poll the same number of votes, $n$ each,
in an election. The counting of votes is usually done in some arbitrary order and
therefore, during the counting process A may lead for some period and B may lead for
some period. Prove that the number of ways of counting the votes such that A never trails B is
the $n^{th}$ catalan number.
I know that in such questions we are supposed to analyse the problem by dropping a few terms. For example, in this one, we can drop the $2n^{th}$ vote and check it for $2n-1$ votes, but it doesn't seem to be correct since I won't get catalan number. Can someone please help in getting a combinatorial argument/proof for this? Thanks!
combinatorics recurrence-relations catalan-numbers
asked Nov 21 '18 at 6:04
Ankit Kumar
1
1
Read sections 1.3 and 2.4 of geometer.org/mathcircles/catalan.pdf
– Robert Z
Nov 21 '18 at 6:35
Ya it's nice and thanks a lot for it. But, i was looking for more of a combinatorial proof
– Ankit Kumar
Nov 21 '18 at 7:04
1
Consider each vote for $A$ as an opening parenthesis, and each vote for $B$ as a closing parenthesis. What are you counting now?
– Jean-Claude Arbaut
Nov 21 '18 at 7:45
add a comment |
Suppose two candidates A and B poll the same number of votes, $n$ each,
in an election. The counting of votes is usually done in some arbitrary order and
therefore, during the counting process A may lead for some period and B may lead for
some period. Prove that the number of ways of counting the votes such that A never trails B is
the $n^{th}$ catalan number.
I know that in such questions we are supposed to analyse the problem by dropping a few terms. For example, in this one, we can drop the $2n^{th}$ vote and check it for $2n-1$ votes, but it doesn't seem to be correct since I won't get catalan number. Can someone please help in getting a combinatorial argument/proof for this? Thanks!
combinatorics recurrence-relations catalan-numbers
asked Nov 21 '18 at 6:04
Ankit Kumar
1
Suppose two candidates A and B poll the same number of votes, $n$ each,
in an election. The counting of votes is usually done in some arbitrary order and
therefore, during the counting process A may lead for some period and B may lead for
some period. Prove that the number of ways of counting the votes such that A never trails B is
the $n^{th}$ catalan number.
I know that in such questions we are supposed to analyse the problem by dropping a few terms. For example, in this one, we can drop the $2n^{th}$ vote and check it for $2n-1$ votes, but it doesn't seem to be correct since I won't get catalan number. Can someone please help in getting a combinatorial argument/proof for this? Thanks!
combinatorics recurrence-relations catalan-numbers
combinatorics recurrence-relations catalan-numbers
asked Nov 21 '18 at 6:04
Ankit Kumar
1
asked Nov 21 '18 at 6:04
Ankit Kumar
1
edited Nov 21 '18 at 7:41
asked Nov 21 '18 at 6:04
Ankit Kumar
1
asked Nov 21 '18 at 6:04
Ankit Kumar
1
asked Nov 21 '18 at 6:04
Ankit Kumar
1
1
1
Read sections 1.3 and 2.4 of geometer.org/mathcircles/catalan.pdf
– Robert Z
Nov 21 '18 at 6:35
Ya it's nice and thanks a lot for it. But, i was looking for more of a combinatorial proof
– Ankit Kumar
Nov 21 '18 at 7:04
1
Consider each vote for $A$ as an opening parenthesis, and each vote for $B$ as a closing parenthesis. What are you counting now?
– Jean-Claude Arbaut
Nov 21 '18 at 7:45
add a comment |
1
Read sections 1.3 and 2.4 of geometer.org/mathcircles/catalan.pdf
– Robert Z
Nov 21 '18 at 6:35
Ya it's nice and thanks a lot for it. But, i was looking for more of a combinatorial proof
– Ankit Kumar
Nov 21 '18 at 7:04
1
Consider each vote for $A$ as an opening parenthesis, and each vote for $B$ as a closing parenthesis. What are you counting now?
– Jean-Claude Arbaut
Nov 21 '18 at 7:45
1
1
Read sections 1.3 and 2.4 of geometer.org/mathcircles/catalan.pdf
– Robert Z
Nov 21 '18 at 6:35
Read sections 1.3 and 2.4 of geometer.org/mathcircles/catalan.pdf
– Robert Z
Nov 21 '18 at 6:35
Ya it's nice and thanks a lot for it. But, i was looking for more of a combinatorial proof
– Ankit Kumar
Nov 21 '18 at 7:04
Ya it's nice and thanks a lot for it. But, i was looking for more of a combinatorial proof
– Ankit Kumar
Nov 21 '18 at 7:04
1
1
Consider each vote for $A$ as an opening parenthesis, and each vote for $B$ as a closing parenthesis. What are you counting now?
– Jean-Claude Arbaut
Nov 21 '18 at 7:45
Consider each vote for $A$ as an opening parenthesis, and each vote for $B$ as a closing parenthesis. What are you counting now?
– Jean-Claude Arbaut
Nov 21 '18 at 7:45
add a comment |
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1
Read sections 1.3 and 2.4 of geometer.org/mathcircles/catalan.pdf
– Robert Z
Nov 21 '18 at 6:35
Ya it's nice and thanks a lot for it. But, i was looking for more of a combinatorial proof
– Ankit Kumar
Nov 21 '18 at 7:04
1
Consider each vote for $A$ as an opening parenthesis, and each vote for $B$ as a closing parenthesis. What are you counting now?
– Jean-Claude Arbaut
Nov 21 '18 at 7:45