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Default Count combinations

I didn't conclude that he wanted the most frequent total. I thought he
wanted to be able to vary all 3 parameters -- number range, size of set, and
desired total. But you may be correct.

"Dana DeLouis" wrote in message
...
I don't have an answer, just an observation...
With the range of numbers 1-10 (Instead of 1-49), and you take all
combinations of 6:
The smallest combination total is 21 (1,2,3,4,5,6), and occurs only once.
The Largest combination total 45 {5, 6, 7, 8, 9, 10}, and also occurs

once.

The average of these two numbers (21+45)/2 is the number that occurs the
most often. The answer here is 33.
In a range of 10 numbers taken 6 at a time, a total of 33 occurs the most.

If the Range is 1-49, taken 6 at a time, the total that occurs the most
often is:
(21 + 279)/2

150

It appears the op is asking about the total count of the sum that occurs

the
most often.

The brute force method generates many combinations. Instead, I looked at
the total of combinations from a smaller range of numbers.
First, Range 1-6, and made permutations of 6 (basically just one solution)
then Range 1-7, Range 1-8, etc
I did this quickly for the range 1-(6-20)
The number that occurs the most (when grouped as 6 numbers) generated the
following sequence.

{1, 1, 4, 8, 18, 32, 58, 94, 151, 227, 338, 480, 676, 920, 1242, 1636}

Apparently, this is a know integer sequence (A001977). The bad news is
that it does not give the generating function directly. It is listed as a
form of "Restricted partitions."


http://www.research.att.com/cgi-bin/...i?Anum=A001977

Maybe someone with more knowledge can shed some light on this. I have

never
seen a result listed like this before, so this would be new to me.

Harlan?
Tushar?

The program Mathematica has the function PartitionsP, which I am sure your
Maple program does also:

Information[PartitionsP]

"PartitionsP[n] gives the number p(n) of unrestricted partitions of the
integer n."
Attributes[PartitionsP] = {Listable, Protected}

Here are the first few terms...
Table[PartitionsP[n], {n, 0, 15}]

{1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176}

It looks like this function fits in somewhere, but I do not know enough
about it. I don't see any pattern.

Harlan? Tushar?
--
Dana DeLouis
Using Windows XP & Office XP
= = = = = = = = = = = = = = = = =


"TwIsTeEeR" wrote in message
able.rogers.com...
From a set of integers from 1 to 49,
take 6 unique integers,
that if we sum them up, this sum is equal to 150.

Conditions:
1. The set numbers are integers from 1 to 49
2. subset size is 6
3. sum of the selected subset numbers is 150

My questions:
A. How many sets (combinations) of 6 unique numbers exist that their sum
is 150?
B. Do you know of a function that can calculate that quantity
of combinations, for different values of conditions (1), (2), & (3)?

Thank you,








 
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