Probability and Statistical Inference. Robert Bartoszynski

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      Multiplying the numerator and denominator by we get

      (3.11)

      In this section, we tacitly assume that is an integer with .

      Observe also that the symbol makes sense for and , in view of the convention that . Thus, we have

(3.12)

for all integers and such that . For , we have , since there is only one empty set, and since only one set of size can be selected out of a set of size . Formula (3.12) gives correct values, namely

      (3.13)

      We will now study some properties of the binomial coefficients . First,

(3.14)

which follows from the symmetry in formula (3.10). One can also prove (3.14) by observing that choosing a set of size is equivalent to “leaving out” a set of size . The number of different sets of size chosen must be equal to the number of different sets of size “chosen” by leaving them out.

      We will now prove the following theorem:

Theorem 3.3.2 (Pascal's Triangle) The binomial coefficients satisfy the relation

(3.15)

Proof: The formula can be easily proved by expressing the left‐hand side using (3.8) and reducing it algebraically to get the right‐hand side. It is, however, more instructive to rely on the interpretation of the coefficients as СКАЧАТЬ