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visits member for 2 years, 11 months
seen Apr 7 at 6:27

Jan
29
awarded  Notable Question
May
24
awarded  Popular Question
May
1
awarded  Yearling
Jan
8
asked Can a Hermitian operator on a tensor product space be represented as a sum of tensor products of Hermitian operators?
Jan
3
awarded  Good Answer
Jan
2
awarded  Nice Answer
Jan
1
awarded  Yearling
Jan
1
answered Are there real-life relations which are symmetric and reflexive but not transitive?
Aug
5
comment What is the sum of only half the exponential terms that give the Dirac comb?
Got it - thanks!
Aug
5
accepted What is the sum of only half the exponential terms that give the Dirac comb?
Aug
5
asked What is the sum of only half the exponential terms that give the Dirac comb?
Jun
8
awarded  Caucus
Mar
7
accepted How many ways can $n$ adults, $k_1$ boys and $k_2$ girls be seated in a line such that no two children of the same sex sit next to each other?
Mar
7
comment How many ways can $n$ adults, $k_1$ boys and $k_2$ girls be seated in a line such that no two children of the same sex sit next to each other?
@dtldarek: why don't you edit your answer to include joriki's correction, and I'll accept it?
Mar
7
comment How many ways can $n$ adults, $k_1$ boys and $k_2$ girls be seated in a line such that no two children of the same sex sit next to each other?
As @joriki pointed out, I did have a mistake in my first comment. The expression I gave for $X_m$ is incorrect, but I believe joriki's expression is correct.
Mar
7
comment How many ways can $n$ adults, $k_1$ boys and $k_2$ girls be seated in a line such that no two children of the same sex sit next to each other?
Nice! This is a similar approach to @dtldarek's answer, only simpler. Though, after joriki's correction that answer becomes simple as well.
Mar
7
comment How many ways can $n$ adults, $k_1$ boys and $k_2$ girls be seated in a line such that no two children of the same sex sit next to each other?
But I think you can put a boy in the left most place in line. If there is an adult there he is next to the adult, and if there is no adult then there is a pair, and he is next to a girl. In both cases it is possible. Also, I don't see how a binomial can ever be equal to zero, since it is defined as a positive integer.
Mar
7
comment How many ways can $n$ adults, $k_1$ boys and $k_2$ girls be seated in a line such that no two children of the same sex sit next to each other?
This approach is a great idea! I think your formula needs some corrections though. First of all, swapping $a$ and $b$ shouldn't matter, so we can assume WLOG $a \ge b$. Second of all, since we can always put a boy in the left most place in the line (whether there's an adult there or not), I think the expression for $X_m$ is simply: $\binom{n+1}{a-m}(n+m)!$. Third, It's not hard to see that there is a minimal number of pairs which might be greater than zero, so the summation should be $\sum_{m=\max{(0,a-n-1)}}^b$.
Mar
7
comment How many ways can $n$ adults, $k_1$ boys and $k_2$ girls be seated in a line such that no two children of the same sex sit next to each other?
Naturally the initial condition of the recurrence is $f(n,k,1)=f(n,1,k)=\binom{n+2}{k}(n+1)!k!$, since this is like having only boys or only girls.
Mar
7
comment How many ways can $n$ adults, $k_1$ boys and $k_2$ girls be seated in a line such that no two children of the same sex sit next to each other?
I managed to derive a recurrence relation for $f(n,k_1,k_2)$. There are two ways to seat the last girl: a. first seat everyone except the last girl ($f(n,k_1,k_2-1)$ configurations), and then we have $n+k_1-k_2+2$ places to seat the last girl; b. choose two boys ($k_1(k_1-1)$ possible choices, their order matters), seat only them next to each other (it's like having one less boy, so $f(n,k_1-1,k_2-1)$ configurations), and then put the last girl between them. Putting it together we get: $f(n,k_1,k_2)=(n+k_1-k_2+2)f(n,k_1,k_2-1)+k_1(k_1-1)f(n,k_1-1,k_2-1)$.