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The branch of mathematics which studies the number of different ways of
arranging things is called combinatorial analysis. We need to know
how many different ways there are of arranging objects which are made up
of two groups of and indistinguishable objects. This is a
pretty tough problem! Let us try something a little easier to begin with.
How many ways are there of arranging distinguishable
objects? For instance, suppose that we have six pool balls, numbered one
through six, and we pot one each into every one of
the six pockets of a pool table (that is, topleft, topright, middleleft,
middleright, bottomleft, and bottomright). How many different ways
are there of doing this? Well, let us start with the topleft pocket.
We could pot any one of the six balls into this pocket, so there are
6 possibilities. For the topright pocket we only have 5 possibilities,
because we have already potted a ball into the topleft pocket, and it
cannot be in two pockets simultaneously. So, our 6 original possibilities
combined with these 5 new possibilities gives ways of potting
two balls into the top two pockets. For the middleleft pocket we have
4 possibilities, because we have already potted two balls. These possibilities
combined with our possibilities gives
ways
of potting three balls into three pockets.
At this stage, it should be clear that
the final answer is going to be
.
Well,
is a bit of a mouthful,
so to prevent us having to say (or write) things like this, mathematicians have invented
a special function called a factorial. The factorial of a general
positive
integer is defined

(17) 
So, , and
, and
, and so on. Clearly, the
number of ways of potting six pool balls into six pockets is (which
incidentally equals 720). Since there is nothing special about pool balls, or
the number six, we can safely infer that the number of different ways of
arranging distinguishable objects, denoted , is given by

(18) 
Suppose that we take the number four ball off the pool
table and replace it
by a second number five ball. How many different ways are there of potting
the balls now? Well, consider a previous arrangement in which the number five
ball was potted into the topleft pocket and the number four ball was potted
into the topright pocket, and then consider a second arrangement which only
differs from the first because the number four and five balls have been
swapped around.
These arrangements are now
indistinguishable, and are therefore counted as a single arrangement, whereas
previously they were counted as two separate arrangements. Clearly, the
previous arrangements can be divided into two groups, containing equal numbers
of arrangements, which differ only by the permutation of the
number four and five balls. Since these balls are now indistinguishable, we
conclude that there are only half as many different arrangements as
there were before. If we take the number three ball off the table and replace
it by a third number five ball, we can split the original arrangements into
six equal groups of arrangements which differ only by the permutation
of the number three, four, and five balls. There are six groups because there
are separate permutations of these three balls. Since the number three,
four, and five balls are now indistinguishable, we conclude that there are
only the number of original arrangements. Generalizing this result, we
conclude that the number of arrangements of indistinguishable and
distinguishable objects is

(19) 
We can see that if all the balls on the table are replaced by number
five balls then there is only possible arrangement. This
corresponds, of course, to a number five ball in each pocket.
A further straightforward generalization tells us that the
number of arrangements of two groups of and indistinguishable
objects is

(20) 
Next: The binomial distribution
Up: Probability theory
Previous: The twostate system
Richard Fitzpatrick
20060202