Derivation of the probability distribution function of a binomial distribution

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misogynisticfeminist
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Is there a way to derive

[tex]P (X=r) =^nC_r p^r q^{n-r} , r= 0, 1, 2,..., n[/tex]

where [tex]X: B(n,p)[/tex]

where n is the total number of bernoulli experiments,

p the probability of success

q, the probability of failure.
 
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Yes, just think about it, it's just simple combinatorics (you omitted to mention independent trials, which is, I'm sure important), and no harder than working out how to choose r from n (in fact it is the same).
 
This is not a problem as matt goes into. Suppose we have XXX and YY,then how many ways can combinations occur? Well there are five elements in 5! ways, but 3 of them are similar and the other two are similar, so it's 5!/(3!2!)=10 distinct ways. YYXXX, YXYXX, YXXYX, YXXXY, XYXXY, XXYXY, XXXYY, XYYXX, XXYYX, XYXYX.