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Corollary: Let n be a positive integer, and r an
integer in the range 0 <= r <= n.
Then P(n,r) = n!/(n-r)!
Proof: For r in the range 1 <= r <= n, this follows
from the previous theorem and the fact that
n!/(n-r)! = n(n-1) ... (n-r+1).
For r=0, we have P(n,0)=1, which equals n!/(n-0)!
26 Permutation Example How many permutations of the letters ABCDEFGH contain the
Let us regard ABC, D, E, F, G, and H as blocks. Any permutation
of these six blocks will yield a valid permutation containing ABC,
and there are no others. Therefore, we have 6!=720
permutations of the letters ABCDEFGH that contain ABC as a
block. 27 Combinations
Let S be a set of n elements. An r-combination of
S is a subset of r elements from S.
The number of r-combinations of a set S with n
elements is denoted by
r 28 Number of r-Combinations
Theorem: The number of r-combinations of a set
with n elements is given by
(n − r)!r!
Proof. We can form all r-permutations of a set
with n elements by first choosing an rcombination and then ordering the r elements in
all possible ways. Thus, P(n,r)=C(n,r)P(r,r).
Since (r-r)!=0!=1, this yields our claim. Binomial Coefficient Identity
Corollary: We have n
Proof: Let S be a set with n elements. Each subset A of S is
determined by its complement Ac, which specifies the elements of S
that are not contained in A. Therefore, we can use double counting:
The number C(n,r) of subsets of cardinality r of S corresponds to
the number of “complements of subsets of cardinality r in S”. Since
|A|=r iff |Ac|=n-r, the complements of subsets of cardinality r of S
correspond to subsets of cardinality n-r of S.
Thus, C(n,r) = # of r-subsets of S
= # of complements of r-subsets of S = C(n,n-r), as claimed.
30 Counting Subset Identity
Theorem: For any nonnegative integer n, we have
k n =2 Proof: Let S be a set...
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This note was uploaded on 03/24/2014 for the course CSCE 222 taught by Professor Math during the Fall '11 term at Texas A&M.
- Fall '11