Matthew Aldridge
@mpaldridge.bsky.social
But blue. https://mpaldridge.github.io
I like to log on every morning and spend hours reading the thoughts of the madding crowd.
November 26, 2025 at 8:18 AM
I like to log on every morning and spend hours reading the thoughts of the madding crowd.
Hmm… So the Beta function B(a, b) for integer a and b is the reciprocal of the multinomial coefficient “n choose (a-1, b-1, 1)”, where n = a + b – 1. But I’m not sure what to do with that fact…
November 22, 2025 at 11:32 AM
Hmm… So the Beta function B(a, b) for integer a and b is the reciprocal of the multinomial coefficient “n choose (a-1, b-1, 1)”, where n = a + b – 1. But I’m not sure what to do with that fact…
It would essentially be the same:
n^(n falling) / k1! k2! k3!
where n^(n falling) = n!
But if you have n items and want to paint k1 red and k2 blue, you can have
n^((k1 + k2) falling) / k1! k2!
without needing to explicitly write down the k3 = n - k1 - k2 items that are left unpainted.
n^(n falling) / k1! k2! k3!
where n^(n falling) = n!
But if you have n items and want to paint k1 red and k2 blue, you can have
n^((k1 + k2) falling) / k1! k2!
without needing to explicitly write down the k3 = n - k1 - k2 items that are left unpainted.
November 21, 2025 at 9:56 PM
It would essentially be the same:
n^(n falling) / k1! k2! k3!
where n^(n falling) = n!
But if you have n items and want to paint k1 red and k2 blue, you can have
n^((k1 + k2) falling) / k1! k2!
without needing to explicitly write down the k3 = n - k1 - k2 items that are left unpainted.
n^(n falling) / k1! k2! k3!
where n^(n falling) = n!
But if you have n items and want to paint k1 red and k2 blue, you can have
n^((k1 + k2) falling) / k1! k2!
without needing to explicitly write down the k3 = n - k1 - k2 items that are left unpainted.
Ah! But the multinomial coefficient for n choose k1, k2, ..., km is
n^(|k| falling) / k1! k2! … km!
where |k| = k1 + … + km.
And this is even better! Because it works not only when |k| = n, when n^(n falling) = n!, but also when |k| < n, so you don't even have to write down the left-over number!
n^(|k| falling) / k1! k2! … km!
where |k| = k1 + … + km.
And this is even better! Because it works not only when |k| = n, when n^(n falling) = n!, but also when |k| < n, so you don't even have to write down the left-over number!
November 21, 2025 at 9:49 PM
Ah! But the multinomial coefficient for n choose k1, k2, ..., km is
n^(|k| falling) / k1! k2! … km!
where |k| = k1 + … + km.
And this is even better! Because it works not only when |k| = n, when n^(n falling) = n!, but also when |k| < n, so you don't even have to write down the left-over number!
n^(|k| falling) / k1! k2! … km!
where |k| = k1 + … + km.
And this is even better! Because it works not only when |k| = n, when n^(n falling) = n!, but also when |k| < n, so you don't even have to write down the left-over number!
Can you point me somewhere to read about this?
November 19, 2025 at 5:57 PM
Can you point me somewhere to read about this?