how can i find all the different combinations of the numbers 1-6?

Answer 1

This is an interesting problem. I'll assume you want to count how many there are, not list them (there are way too many to want to list!) Sadly, nobody understands this problem, and they're giving you outrageously wrong answers! : (

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So, here's what you do to count:

You could use the formulas:

6C0 + 6C1 + 6C2 + ... + 6C5 + 6C6

to add up all the different combinations. This would give you an answer, assuming you're familiar with binomial coefficients (n choose k). See also:
http://en.wikipedia.org/wiki/Binomial_coefficients

On the other hand, let's say you want an answer that is a little bit more insightful - you can count it up as follows:

Include 1 or not: 2 choices
Include 2 or not: 2 choices
Include 3 or not: 2 choices
....etc
Include 6 or not: 2 choices

So there are 2×2×2×2×2×2 = 2^6 ways to do it. See also:
http://www.theory.csc.uvic.ca/~cos/inf/comb/SubsetInfo.html

This is the reason that the rows of Pascal's triangle add up to powers of two if you add across each row. See:
http://en.wikipedia.org/wiki/Pascal%27s_triangle

So the answer is 2^6 = 64

PLEASE NOTE: picking no numbers IS a possibility, despite what uninformed people say.

If I want to list the games a football team wins, winning no games is a possibility. Now let's say they played 6 games - picking the wins out of 6 games can certainly have "none" as an option.

Say I want to pick some number of cookies out of a jar. Taking zero cookies is possible. 0 is a number of cookies I can take, albeit an unsatisfying amount of cookies.

This is called a trivial combination. It is the "empty set" of combinations, so to speak. Despite seeming unimportant, it is actually highly important (just like the empty set is important to set theory). There is not a single mathematician or good math student would count 2^6-1, because "choosing none" is ALWAYS considered just as valid as any other combination.

Answer 2

For each number 1 thru 6 you can either include it or not. However, I assume you want the combination to include at least one number. So the number of combinations is:

2^6 - 1 - 64 - 1 = 63

The minus one is to exclude the possibility of including no numbers.

Cheeser had the right idea but forgot to exclude the possibility of picking no numbers.

Answer 3

First digit has 6 possibilities, Second digit has 5 possibilities, etc. Answer is 6*5*4*3*2*1 = 720.

Answer 4

the question is still unclear. Usually when we talk about numbers, we deal with permutations.

Permuation is a number of arrangements and order matter.

There are 1,2,3,4,5,6 ; 6 digits from 1 to 6

assume you ask for 6 digits numbers
6 x 5 x 4 x 3 x 2 x 1 = 720 ways

and if the number of digits can be repeated then:
6^6 = 46,656

Answer 5

it is just a matter of permutations and combinations .Say you want to find possible the number of digit numbers that can be made from 1-6 then there are 7 digits and a 3 digit no so we call it 7p3or 7!/(7-3)! and in general for chosing r numbers from n digits the answer is n!/(n-r)! where n!means n(n-1)(n-2)...3.2.1
this feild of mathematics is called combinatorial mathematics .we define0!=1 so in your question we are chosing 7 from 7 so it is 7!=5040 cool na

Answer 6

you have to spend very much time on finding relations between no. 1-6 because their are too many relations as subtraction, addition, multiplication,division,addition of 2 digits ,addition of 3 digits.

Answer 7

6C6=1