Combinations In Probability
The concept of
combinations is easy to understand. A combination
describes picking an unordered subgroup from a larger
group of items. A permutation is similar, but it is used
when the order within the subgroup matters. For example,
imagine that you have three children. At work you were
given free movie passes. This should make your children
very happy, except there is a problem. You were only
given two passes. One of your children must stay at home
while the other two go. The order in which you pick the
lucky two doesn't matter; both will get to see a movie.
This is an example of a combination.
Now imagine that you arrived late to the cinema,
probably because you spent much time consoling your
third child. There are only two movies playing, and each
screening has only one seat left. You decide that the
first child can choose the movie, and the second must
see the other show. Now the order of the two children
matters, and we have converted this example into a
permutation.
Of course, your children want to know how lucky, or
unlucky, they were. They want to know the number of
possible ways that the two moviegoers could have been
picked. Let's assume that some people have left the
theater, and the two kids can see whatever they want. We
are back to a combination problem and need to count the
number of combinations. If the children are Jane, John
and Sue, then these are the possibilities:
Jane, John
Jane, Sue
John, Sue
There are three possible combinations. Whether this
information actually makes the children any happier, is
a question left to the reader.
Although it's fine to count the number of combinations
by drawing out the possibilities in this way, it only
works with small numbers. What if this is the old woman
who lives in a shoe? What if there are nine children and
two movie tickets? It isn't feasible to draw out all the
ways to choose two out of nine. She needs the general
formula: C(n,r) = n! / ( (r)!(n-r)! )
The exclamation point (!) is a factorial, "n" is the
total number of children and "r" is the number of
moviegoers. So, "n" is nine, and "r" is two.
The number of ways to choose two moviegoers out of nine
children is:
C(9,2)
= 9! / ( 2! * (9-2)! )
= 9! / (2! * 7!)
To calculate a factorial, just multiply the factorial
number with all of the positive integers below it.
9! = 9*8*7*6*5*4*3*2*1
2! = 2*1
7! = 7*6*5*4*3*2*1
The number of combinations for the mother who lives in a
shoe is:
(9*8*7*6*5*4*3*2*1) / ( (2*1) * (7*6*5*4*3*2*1) )
= 362,880/(2 * 5040)
= 362,880/10080
= 36
For those with an algebra background, you'll notice that
the formula can be simplified by canceling out terms
between the numerator and denominator. If not, then just
find the calculator on your smartphone! Copyright
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