What is the formula...?

Question

What is the formula I use to calculate all possible combinations in a number sequence? For example, if I want to calculate ALL possible combinations with the numbers "12345", what formula do I use?

Answer 1

5 different numbers, so it's 5! (factorial) = 5*4*3*2*1 = 120 (:

Answer 2

The question is based on a concept called Permutations. The number of possible permutations (what you actually want is permutations & not combinations; combinations is a different concept) of n things taken r at a time is written as nPr, rhose value is n! / (n-r)!.
So, the answer to your question is 5P5 [since you want the no. of permutations of 5 numbers taken all at a time], whose value is 5! / (5-5)! = 5! / 0! = 5! / 0! = 5! / 1 = 5! = 5*4*3*2*1 = 120
[it is a general convention to take 0! = 1].

Answer 3

The term "COMBINATION" in the math world means the possible arrangements of something pulled from the same size or a larger group. (ORDER does NOT matter)

The formula for Combinations is:

n! / [(n-r)!*r!]

n! = n*(n-1)*......*5*4*3*2*1

(There is 0! which by definition is equal to "1", which if you need, I will explain)

In your example of "12345" the anwer is "1". As an example ... you are selling all 5 of 5 books you own. There is one way of doing this.

5!/[(5-5)!*5!] = 5!/(0!*5!) = 5!/(1*5!) = 120/120 = 1

If you want to sell three of the five books, then there is "10" ways to sequence those books. "b1, b2, b3" is the same as "b3, b2, b1", but the other sequences (which are NOT the same) are "b2, b3, b4" and "b3, b4, b5", and "b1, b3, b4" ...

5!/[(5-3)!*3!] = 5!/(2!*3!) = 120/12 = 10

If order is important to you, then you have a PERMUTATION.

n! / (n-r)!

Selling all 5 books, but it matters how you arrange them ... b1, b2, b3, b4, b5 is completely different than b5, b1, b2, b3, b4 ... then there are "120" sequences.

5!/(5-5)! = 5!/0! = 120/1 = 120

Selling three of five books where order matters ... would be "60" sequences. "b1, b2, b3" is completely different than "b3, b2, b1", and they of course are different than "b2, b3, b4", which are different than "b4, b3, b2".

5!/(5-3)! = 5!/2! = 120/2 = 60

If any of this makes sense to you.

Answer 4

I believe it is called factorial, it is use to figure out the number of combinations that a specific number can produce. It is also know as the binomial theorem! the symbol on your calculator is X! .

Answer 5

I don't exactly know, but i pretty sure that multiplying the nos. is not an answer...

For e.g., since 12345 n 67890 each have 5 nos., both should have same no. of combinations possible... but if i multiply them, the give greatly different values...

Answer 6

never go by formula

think u have 5 boxes

now in one box u can put any of the 5 no.
in other u cab only put 4 no. as one have already used
now in third one only three
now in fourth 2 no.
in last only 1

5*4*3*2*1=120

Answer 7

it's 5*4*3*2*1 just countdown and multiply depending on how many numbers there are.