|Feb20-08, 02:30 PM||#1|
the question is:
let k, n, and k1, . . . , kn be given natural numbers, such that
k1 + . . . + kn = k.
Assume that k musicians shall be distributed to n orchestras such that exactly
ki musicians play in the ith orchestra. Prove that there exist exactly
k!/(k1! · · · kn!)
is it possible to use induction to answer this? i can prove it by using the choose function to find all the possible distributions. in that way i get a proof for the statement, but i am unable to assume that it is correct for n, and then show it is correct for n+1. can someone give me some ideas?
|Feb20-08, 03:13 PM||#2|
Please do not cross-post.
|Feb22-08, 11:05 AM||#3|
I've done the same calculus for something else.
|Feb22-08, 11:38 AM||#4|
If you look at derivation of multinomial distribution of probability (look at the coefficients), you get the answer.
|Feb24-08, 08:41 AM||#5|
yeah none of that really helped, but for those who are interested in the proof using induction here it is (it seems kind of redundant since it uses the combinatorial proof within the inductive proof, but anyway):
Induction step: Assume that the assertion is true for some n with n >= 2,
and for all numbers k, k1, . . . kn, such that k1 + . . . + kn = k. Then, let
k, k1, . . . , kn+1, such that k1 + . . . + kn+1 = k. Number the orchestras
from 1 to n+1. We first ”unify” the last two orchestras - with numbers n and
n + 1 - to one orchestra, which we call O. Then distribute the musicians such
that exactly ki of them play in the ith orchestra, (i = 1, . . . , n − 1), and let
the rest (= kn + kn+1 musicians) sit in the orchstra O. By assumption, there
k!/k1! · · · kn−1!(kn + kn+1)!
such distributions. To complete the task, we divide the orchstra O into two
parts - of size kn respectively kn+1. As we have seen in the begin of induction,
there are exactly
(kn + kn+1)!/kn!kn+1!
such divisions. In conclusion, there are exactly
D · D0 =
k!/(k1! · · · kn−1!kn!kn+1!)
distributions of the k musicians such that exactly ki musicians play in the ith
orchestra, (i = 1, . . . , n + 1). In other words, the assertion is true for n + 1.
is it just me or is that kind of pointless. i guess what i mean is, doesn't the combinatorial proof CONTAIN the inductive proof?
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