If you read my post carefully, (assuming it made
your server), you will see that
it, indeed, gives the combinations solution you gave
below, and shows that it is equivalent to
another, more elegant way of arriving at the same
answer, i.e.,
2^N = (in this case) 2^5 = 32
Jim Steiger
On 30 Apr 2001 12:18:55 -0700, [EMAIL PROTECTED] (Robert R
Johnson) wrote:
>Several have written to this thread and I believe there has been some
>misleading information passed along and intermixed with correct
>information.
>
>On Sun, 29 Apr 2001 11:26:11 -0400 "Zina Taran"
><[EMAIL PROTECTED]> writes:
>> I believe, the thrust of the "fries" reply was the overcount in the
>> 5*4*3*..... response rather than an expression of culinary
>> preferences.
>
>Multiplying 5*4*3... is the kind of multiplication used when determining
>the 'number of permutations'.
>
>I seem to remember the question asked for "number of combinations".
>
>The first step to obtaining the answer is to be sure you understand what
>you are 'counting.'
>
>If PERMUTATIONS -
>If you consider a "hamburger with lettuce and mayo" to be different than
>a "hamburger with mayo and lettuce", then you want to count the "number
>of permutations." Permutations involve the concept of "order" [rather it
>be 'the order in which you listed the condiments when placing the order'
>or 'the order in which the condiments are placed on your burger'] as an
>integral part of the problem. As applied to this particular problem,
>there were 5 condiments - for the first condiment selected there are 5
>choices, for the second condiment selected there would be 4 choices
>available, and so on. Then by multiplying, you find the number of
>possibilities. I seriously doubt that you are interested in counting the
>"number of permutations."
>
>If COMBINATIONS -
>If what you want to know is "how many different combinations of
>condiments can be ordered", with no regards to an "order" concept, then
>you are interested in counting the "number of combinations". It is this
>concept that I would think answers the question asked.
>
>One way to approach this is to look at and determine the count for each
>case, then find the total. [This approach was suggested in an earlier
>e-mail.]
>Five (5) condiments to choose from -
>1) find the number of ways you can select exactly zero (0) condiments -
>there is 1,
>2) find the number of ways you can select exactly one (1) condiment -
>there are 5,
>3) find the number of ways you can select exactly two (2) condiments -
>there are 10,
>4) find the number of ways you can select exactly three (3) condiments -
>there are 10,
>5) find the number of ways you can select exactly four (4) condiments -
>there are 5,
>6) find the number of ways you can select exactly five (5) condiments -
>there are 1,
>I'll leave it to you to find each of them.
>Answer: 1+5+10+10+5+1 = 32 different possible combinations of condiments
>can be ordered.
>
>If you are not interested in all the cases individually, there is a
>shorter way. Think of each different condiment that you can order, say
>pickles - you have 2 choices - 'order it' or 'don't order it'. The same
>two (2) options are available for each of the other condiments - catsup,
>mayo, lettuce, etc. [even for the fries and onion rings, if you want them
>included as a condiment on your burger. :-} ]
>
>Now what you have is a special case of the basic "Multiplication Rule".
>Two (2) choices for each condiment, 2*2*2* ... *2, using 'n' 2's, or
>2**n. Thus, if you have 5 condiments to choose from, there are 2**5 =
>32 different combinations of condiments can be ordered for your
>hamburger.
>
>I hope this helps.
>-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>Robert R. & Barbara S. Johnson
>E-mail: [EMAIL PROTECTED]
>Post: 84 West Lake Rd., Branchport, NY 14418
>
>
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