combinations
rob
This is the problem: How many five digit numbers can there be if a
five digit number is such that each digit is either 0,1,2,3,4,5 or 6?
Assume 00000 is the first number and so it goes 00001, 00002, etc.. .
Really appreciate your help here and thanks.
Will_vK
places.
in this case, it is a base7 number system with five places,
giving 7^5
Cheers,
Daniel C. Bastos
rob wrote:
> I need to know the maximum number of combinations?
I think that the proper word here would be ``permutations.''
See, for example, the article
http://www.mathagonyaunt.co.uk/STATISTICS/ESP/Perms_combs.html
which explains about the differences between the concepts.
The problem is asking for how many distinct strings of 5
symbols you can form by using the symbols 0, 1, 2, 3, 4, 5,
6. So, changing the word in your question to permutations,
the answer is ``yes.'' By ``strings'' I mean a chain of
letters, so order matters.
> This is the problem: How many five digit numbers can there be if a
> five digit number is such that each digit is either 0,1,2,3,4,5 or 6?
> Assume 00000 is the first number and so it goes 00001, 00002, etc.. .
Pick a digit from your set of possible digits and place it
in the first position of your 5-digit string. How many
choices are there? There are 7 because you can pick 0, 1, 2,
3, 4, 5, or 6. Now pick a second digit for the second place
of your 5-digit string. How many choices are there? Still 7
because nothing in the problem is keeping you from reusing
the digits more than one; that is, the strings 11111, or
33333 are legal.
If that isn't enough help, consider a simpler example in
details. Consider two letters A, B and C. How many different
strings of 2 letters can we get with them?
We have 3 choices for the first letter of any string, and 3
choices for the second letter of any string. If you list
them, you'd see
AA
AB
AC
BA
BB
BC
CA
CB
CC
One way to convince yourself that you've listed them all is
to think of an arbitrary string. The arbitrary string would
either start with A, B or C, and notice that you have three
cases for each letter. The second letter of your arbitrary
string would be either A, B or C, and notice that that you
have A, B, and C as second letter for each A, B, C as first
letter. There really shouldn't be anything missing.
It turns out that anytime you have n choices of letter for a
m-letter word, you have n*n* ... * n (m times) which is
written as n^m, if order matters. In the ABC case above, we
listed 9 = 3*3 = 3^2 2-letter strings.
Here's an article about the subject
Brian M. Scott
<dba...@yahoo.com.br> wrote in
<news:20060426203645.000070bb@Saturn> in alt.algebra.help:
> In article <jrrv425rravfgl8le...@4ax.com>,
> rob wrote:
>> I need to know the maximum number of combinations?
> I think that the proper word here would be ``permutations.''
Neither is actually correct, since you can use the same item
more than once; however, you're entirely correct in pointing
out (in the bit that I snipped) that order matters here,
just as it does for permutations. (In the version of this
question in which order doesn't matter, we'd be counting
what are known as multisets. That's a significantly harder
problem, with a more complicated answer, the binomial
coefficient C(11,5).)
[...]
Brian
Darrell
<rob> wrote in message news:jrrv425rravfgl8le...@4ax.com...
Fundamental counting principal:
possibilities of first digit * possibilities of next digit * ...
= 7 * 7 * 7 * 7 * 7
...for the same reason there are 1000 possible 3-digit strings using 0-9 as
digits. Don't read too much into the question.
--
Darrell
rob
Thank you.
If I omit the zeros so we use just 1 thru 6 for a 5 digit number, if I
understand you correctly the max. combos would be base 6 to the fifth
power or 6^5, right ?
Stan Brown
> I need to know the maximum number of combinations?
>
> This is the problem: How many five digit numbers can there be if a
> five digit number is such that each digit is either 0,1,2,3,4,5 or 6?
> Assume 00000 is the first number and so it goes 00001, 00002, etc.. .
These are not combinations, as we use the term in math. In
combinations, the order doesn't matter, so 12345 and 42513 are "the
same".
What you're interested in is "permutations", where each arrangement
counts as different. The number of permutations is usually easier to
compute than the number of combinations.
For instance, to compute your count of possible five-digit numbers,
ask yourself how many possibilities there are for the first digit.
There are seven. Now for _each_ possible first digit there are seven
possibilities for the second digit, giving a total of 7^2 or 49
possibilities for the first two digits. For each of those there are
seven possibilities for the third digit, giving 7^3 or 343
possibilities for the first three digits.
Continue in that manner to solve the complete problem.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/
Will_vK
>understand you correctly the max. combos would be base 6 to the fifth
>power or 6^5, right ?
that is correct. :)
Daniel C. Bastos
Brian M. Scott wrote:
Wikipedia seems to use the terms ``permutation with
repetition'' and ``permutation without repetition.''
Similarly for combination. The article is
http://en.wikipedia.org/wiki/Combinations_and_permutations.
But I see that you're right; I'm thinking on the definition
of permutation: a one-to-one and onto function f: S -> S. So
we can't repeat the objects, otherwise we wouldn't respect
the definition. Does that make the terms, that Wikipedia is
using, inappropriate? It seems so because ``a permutation
with repetition'' seems to be a contradiction.
Brian M. Scott
<dba...@yahoo.com.br> wrote in
<news:20060427010631.00005ea4@Saturn> in alt.algebra.help:
Yes. I'm a bit surprised that no one's caught that and
edited the page in the last year. If I remember (and can
find the time), I may do it myself.
> It seems so because ``a permutation
> with repetition'' seems to be a contradiction.
I agree.
Brian
Daniel C. Bastos
Brian M. Scott wrote:
I described the problem in the discussion page. I'll leave
the changes for you or someone who knows how to fix it
properly. There may be people ``watching'' that page, so
they'll probably notice my item there and will do something
about it.
Stan Brown
<b.s...@csuohio.edu>:
> On Thu, 27 Apr 2006 01:06:31 -0400, "Daniel C. Bastos"
> <dba...@yahoo.com.br> wrote in
> <news:20060427010631.00005ea4@Saturn> in alt.algebra.help:
> > of permutation: a one-to-one and onto function f: S -> S. So
> > we can't repeat the objects, otherwise we wouldn't respect
> > the definition. Does that make the terms, that Wikipedia is
> > using, inappropriate?
>
> Yes. I'm a bit surprised that no one's caught that and
> edited the page in the last year. If I remember (and can
> find the time), I may do it myself.
>
> > It seems so because ``a permutation
> > with repetition'' seems to be a contradiction.
>
> I agree.
I see I too was wrong about the perm "permutations". I thought it and
"combinations" implied "without replacement" by default, but could be
qualified by "with replacement".
What _is_ the proper term for what I've been calling "permutations
with replacement"?
Brian M. Scott
<the_sta...@fastmail.fm> wrote in
<news:MPG.1eba5f8bd...@news.individual.net> in
alt.algebra.help:
[...]
> What _is_ the proper term for what I've been calling
> "permutations with replacement"?
They're simply (finite) sequences. They're also sometimes
called lists; the (very good) discrete math text by
Scheinerman does this.
Brian
rob
Just want to follow up here and say thanks to you guys who replied to
my post above...... thanks to Will, Daniel, Brian, Darrell and Stan.
I really appreciate your help a lot guys !!!
I'm sure I'll be back around August or Septemeber when my next child
enters college and probably has to take college algebra <grin>.
Thanks again guys.