I rediscovered a 'trick' today that I had found some time ago and I thought I might share it with you guys:
In the expansion of (a<sub>1</sub> + a<sub>2</sub> + ... + a<sub>n</sub>)<sup>P</sup> the coefficient of a<sub>1</sub><sup>p<sub>1</sub></sup>a<sub>2</sub><sup>p<sub>1</sub></sup>...a<sub>n</sub><sup>p<sub>n</sub></sup> is equal to P!/p<sub>1</sub>!p<sub>2</sub>!...p<sub>n</sub>! where (p<sub>1</sub> + p<sub>2</sub> + ... p<sub>n</sub> = P)
The basic logic of this is that if you consider that you have p<sub>1</sub> lots of a<sub>1</sub>, p<sub>2</sub> lots of a<sub>2</sub> ... and so on then there are P!/p<sub>1</sub>!p<sub>2</sub>!...p<sub>n</sub>! ways to arrange these elements in a line - where their position in the line corresponds to the order in which they have been 'chosen' from the bracket.
What I find this helps with is if you want to extend the logic of binomial probability to a situation that can be expressed as an n-nomial, i.e. (a<sub>1</sub> + a<sub>2</sub> + ... + a<sub>n</sub>)<sup>P</sup> where a<sub>1</sub> + a<sub>2</sub> + ... + a<sub>n</sub> = 1. An example is: A fair die is thrown 6 times. Find the probability that the six scores obtained will consist of exactly two 6's and four odd numbers (give it a go).
Anyhow, there are common sense alternatives to using this method but it's helped me out a few times so I figured you people might find some uses of your own. Also I've attached an alternative explanation (the first I came up with) which I made previously to send to people rather than explaining it again. Don't bother opening it unless you're interested in another way of explaining it.
In the expansion of (a<sub>1</sub> + a<sub>2</sub> + ... + a<sub>n</sub>)<sup>P</sup> the coefficient of a<sub>1</sub><sup>p<sub>1</sub></sup>a<sub>2</sub><sup>p<sub>1</sub></sup>...a<sub>n</sub><sup>p<sub>n</sub></sup> is equal to P!/p<sub>1</sub>!p<sub>2</sub>!...p<sub>n</sub>! where (p<sub>1</sub> + p<sub>2</sub> + ... p<sub>n</sub> = P)
The basic logic of this is that if you consider that you have p<sub>1</sub> lots of a<sub>1</sub>, p<sub>2</sub> lots of a<sub>2</sub> ... and so on then there are P!/p<sub>1</sub>!p<sub>2</sub>!...p<sub>n</sub>! ways to arrange these elements in a line - where their position in the line corresponds to the order in which they have been 'chosen' from the bracket.
What I find this helps with is if you want to extend the logic of binomial probability to a situation that can be expressed as an n-nomial, i.e. (a<sub>1</sub> + a<sub>2</sub> + ... + a<sub>n</sub>)<sup>P</sup> where a<sub>1</sub> + a<sub>2</sub> + ... + a<sub>n</sub> = 1. An example is: A fair die is thrown 6 times. Find the probability that the six scores obtained will consist of exactly two 6's and four odd numbers (give it a go).
Anyhow, there are common sense alternatives to using this method but it's helped me out a few times so I figured you people might find some uses of your own. Also I've attached an alternative explanation (the first I came up with) which I made previously to send to people rather than explaining it again. Don't bother opening it unless you're interested in another way of explaining it.
).
