On Sat, 28 Apr 2001 20:48:03 +0800, Abdul Rahman <[EMAIL PROTECTED]>
wrote:
>Please help me with my statistics.
>
>Question:
>
>If you order a burger from McDonald's you have a choice of the following
>condiments:ketchup, mustard , lettuce. pickles, and mayonnaise. A
>customer can ask for all thesecondiments or any subset of them when he
>or she orders a burger. How many different combinations of condiments
>can be ordered? No condiment at all conts as one combination.
>
>
>Your help is badly needed
>
>Just an Idiot@leftover
This problem can be solved in two rather different, but equivalent
ways. The fact that they ARE equivalent reveals a well-known, but
surprising identity in combinatorics.
One approach is as follows:
To make the problem simpler, call the condiments A,B,C,D,E
Ask, how many distinctly different ways you can construct burgers
with NO condiments. [There is exactly one such way.]
In combinatorial theory, you would call this "5 choose 0" the number
of combinations of 5 objects taken 0 at a time.
Next, ask, how many ways you can construct a burger with ONE
condiment. [There are 5 such ways] In combinatorial theory,
you would call this "5 choose 1".
You proceed onward, enumerating all such ways.
It is, of course, the sum, from i=0 to 5, of "5 choose i".
This is the brute force method, and rather unrewarding and tedious.
The other method is much simpler, and involves a straightforward
insight. That is, any combination of condiments may be CODED
with a binary digit code. Specifically, if a condiment is ON the
burger, code it 1. If it is not on the burger, code it zero. So the
burger with NO condiments would have the code
00000
Some other examples
A B C D E
-------------------
0 0 0 0 0 (No condiments)
1 0 0 0 0 (Only condiment A)
0 1 0 0 0 (Only condiment B)
1 1 0 0 0 (Condiments A and B)
1 1 1 1 1 (All 5 condiments)
So you see, the answer to your problem can also be thought of in
another way. That is, HOW MANY distinct 5 digit binary codes
can you construct? Each unique code corresponds to a condiment
selection.
The fact that these two distinctly different approaches both yield
correct answers proves a classic result, i.e.
N
Sum (N choose i) = 2^N
i=0
Study what I have written carefully, and you should be able to arrive
easily at the solution to your specific problem, and also understand
the solution to the famous question.
"How many distinct sets [including the null set] can be formed
from N objects."
Hope this helps.
--Jim Steiger
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