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### Combinations of Coins

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Date: 1/28/96 at 20:50:9
From: Anonymous

If I had 6 coins in my pocket, how many combinations can I have?
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Date: 1/31/96 at 17:40:34
From: Doctor Syd
Subject: Re: math problem about coins

Hello!

I'm glad you wrote.  There are a couple of different ways to
approach your problem, and there are also a couple of ways to
interpret it. So, to avoid confusion, let's talk about
6 coins in your pocket. First, you could ask how many combinations
can you have so that that the total amount of money you have is
different for each combination. Or, second, you could ask simply
how many combinations you can have, regardless of what the sum of
the values of the coins is.

The second question is easier and more general, and that is the
one I will use.  Maybe you can try yourself to work out the first
question!

Now, let's assume that there are 5 different coins (the penny, the
nickel, the dime, the quarter, and the half-dollar).  So, how many
different combinations of these 5 types of things can we have if
we want to end up with 6 different objects?  We could ask this
question about lots of different objects...not just coins.  It is
the same as the question, how many ways can we arrange 5 different
types of flowers in a bunch if we want our bunch to contain 6
flowers, and repetition of flowers is allowed?  You can make up
lots of different situations where this method will work!

Okay, so how do we solve it?  One way to solve it is pretty slick
because the same method can be used for other problems that are
very complicated... using this method we set up something called a
generating function that is related to the problem.  If you are

A simpler approach to the problem is the following:  Since we are
looking at combinations (not permuatations), order of the coins is
unimportant.  So, for our purposes, having 5 nickels and a dime is
no different from having a dime and 5 nickels, right?  Let us
impose our own order on the problem and count the number of ways
we can have 6 coins when pennies come first, then nickels, then
dimes, then quarters, and finally half-dollars.  This may seem to
contradict what I just said - that order is unimportant.  But, if
you think about it for a little bit it may become clear.  If we
count the number of combinations with the order described above,
we will have gotten all of the combinations possible.  For
example, the combination of 3 nickels, a quarter, and 2 dimes will
appear in our ordered combination as 3 nickels, 2 dimes, and one
quarter.  Conversely, every combination can be represented by one
of our ordered combinations.

Okay, so now think about sitting at a cash drawer and picking out
6 coins so that you begin with coins of the smallest value and end
with the coins of the biggest value.  You will make 4 different
"changes":  1st you will change from picking pennies to picking
nickels; 2nd you will change from picking nickels to picking
dimes; 3rd you will change from picking dimes to picking quarters;
and fourth you will change from picking quarters to picking half-
dollars.  Now, it may happen that in between the changes  you will
not choose any coins...thus, if you don't choose anything between
the 1st and 2nd change, you have, in essence, not chosen any
nickels.  Also, it may happen that you don't even pick any
pennies.  In this case the first change comes before you even
start picking coins.  It is easiest to understand this concept
with a
diagram:

_ | _ _ | _ || _ _

Here the horizontal lines stand for coins, and the vertical lines
stand for the changes...Thus the above diagram represents the
combination consisting of one penny (that's the first horiz.
line), 2 nickels (those are after the first vert. line), one dime,
and 2 half-dollars.  The above diagram suggests that the number of
combinations of coins will be the number of different ways we can
combine 4 vertical lines and 6 horizontal lines in the above
diagram.  Think of each line (both horiz. and vert. lines) as a
place holder of sorts.  There are a total of 10 places.  Let's
label them 1, 2, 3, ..., 10.  Now from the diagram we can see the
each arrangement of lines is determined by where the four vertical
lines are.  In other words, if we know which places the vertical
lines are in, we can tell where the horizontal lines will be...in
the empty spaces!!  So, in how many ways can we place the 4 lines
in the 10 spots?  Mathematicians who have studied combinatorics
(how you count the number of ways to do things), will spout out
that this is just "10 choose 4."  This is written,
(10)
( 4) (sort of!  Notation with the computer is necessarily
clumsy, but the symbol basically looks like a fraction without the
dividing bar, and with parentheses enclosing it.)

and is equal to 10!/4!6!.

I don't know if you are familiar with this concept of "n choose
m."  Basically, "10 choose 4" is the number of ways to choose 4
elements from a set of 10 elements.  It is equivalent to our
problem, since the number of ways to put the 4 lines in the 10
spaces is the number of ways to choose 4 of the 10 spaces in which
to put the vertical lines.  I won't go into the reasons for the
formula for "10 choose 4" now, but if you are curious, feel free
to write back.

So, what all this tells us is that the number of ways to put the 4
vertical lines in the diagram is 10 choose 4.  Thus, the number of
ways to choose 6 coins is also 10 choose 4.  A pretty surprising
answer, if you've not seen it before!!

I hope this helps some.  If you are confused by anything I said,
please do feel free to write back.

-Doctor Syd,  The Math Forum

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Associated Topics:
Middle School Factorials

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