Page 13 - 3_Dinda Erliananda_Counting Principles

Dinda Erliananda Teaching Materials – SMA Class XII – Enumeration Rules

there are 7 flags, two or more of them are the same color, but we cannot distinguish one

position from another. We know that the flags can be arranged in a permutation of 7!,
but with several flags of the same color, we cannot fully distinguish the permutations.

In the following, we will discuss how to find multiple permutations with some of
the same elements.

If the set of letters { } is permuted by 4, there are not too many

permutations, i.e. = 4! = 24. The results of the permutations are:

abcd back cabd dabc

abdc badc cadb dacb
acbd bcad cbad dbac

acdb bcda cbda dbca

adbc bdac cdab dcab
adcb bdca cdba dcba

If letters and letters are and , then the set of letters is

{ }. The result of the permutation of the four letters with the number of letters

p is 2 and the number of letters q is 2 is that
there are 6 kinds of permutations. So that,

of the elements {p, p, q, q} is 24,

of the elements {p, p} is 2, and

of the elements {q, q} is 2.

So, the number of permutations four by four of the elements {p, p. 4. q) is 12 = 6.
In general, permutations with the same elements can be formulated as follows:

If there is n object with is the first kind, is the second kind, ... and is the k-th

type; With n objects, there are permutations. If P is many different permutations, the
first type has ,the second type has etc. According to the rules, the

permutations are obtained:

Since the number of objects is n elements, then:

so that

where:
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