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Unformatted text preview: Proof. Let S be a set with n elements. Order the
elements of this set.
We can specify an rmultisubset T of S by specifying the multiplicity of the ﬁrst element, the multiplicity of the second element, and so on.
43 Proof (continued)
Let us use n bins and mi balls in the ith bin to
specify that the ith element of S occurs with multiplicity mi in T . Then r balls distributed in the n
bins determine an rmultisubset of S .
Therefore, it suﬃces to count the number of ways
we can distribute r balls into n bins.
This number corresponds to the number of strings
consisting of r bullets • (representing the balls), and
n − 1 bars  (representing the walls between the bins.
Since the positions of the r balls among the n − 1 + r
balls and bars suﬃce to
specifysuch a string, we can
n−1+r
conclude that there are
strings consisting of
r
n − 1 bars and r balls. By construction, this coincides with the number multisubsets with r elements
of a set with n elements, so the theorem is proved. Example
Let S = {a, b, c, d} be a set with four elements.
Then the number of 3combinations with repetitions
from S is given by
4
4+3−1
=
= 20
3
3 45 Permutations with Repetition
Let S be a set with n elements. An rpermutation with
repetition of S is an arrangement of r elements of S
with repetitions allowed. Theorem: The number of rpermutations with
repetition of a set with n elements is given by nr.
Proof. This follows directly from the product rule. 46 Permutations under Indistinguishability Constraints 47 Motivation
So far, we assumed that (a) the elements are clearly
distinguishable and (b) each element is chosen at most once
for a permutation and combinations or (b’) that each
element can be chosen repeatedly in permutations and
combinations.
Now we will allow that (a’) the elements from which we are
allowed to choose are not necessarily distinguishable.
[For example, we could choose from a multiset of m red
balls and n blue balls.] 48 Permutations with Indistinguishable Objects In how many ways can we reorder SUCCESS?
We can choose 3 locations for the Ss in C(7,3) ways.
Then we can choose 1 location for the U in C(4,1) way.
Then we can choose 2 locations for the Cs in C(3,2) ways.
Finally, we can choose 1 location for the E in C(1,1) way.
C(7,3)C(4,1)C(3,2)C(1,1)= 7! 4! 3! 1!
3!4! 1!3! 2!1! 1!0! 49 = 7!
3!2!1!1! = 420 Permuting Indistinguishable Objects
Theorem. Suppose we have n diﬀerent objects that
are of k diﬀerent types, where the objects of the same
type are indistinguishable. If there are ni indistinguishable objects of type i for 1 ≤ i ≤ k , then the
total number of permutations is
n!
.
n 1 ! n 2 ! · · · nk ! 50...
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 Fall '11
 math

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