number combinations 1-80?

Are you into base system 80? (since you constraint was final 10 digit number.)

Then it is 80^10.

If you presume that number 10-80 to be counted as a 2 digit number part of the 10 digit then it gets a bit hairy.

Number combinations 1-80?

combination formula = (n!)/(r!(n-r)!)

let n = 80 and r = 10

= (80!)/(10!(80-10)!)

= (80!)/(10!(70!))

cancel 70!

= ( 80*79*78*77*76*75*74*73*72*71 ) / (10*9*8*7*6*5*4*3*2*1)

= (5974790569203456000) / (3628800)

= 1646492110120

So in conclusion there is 1,646,492,110,120 ways to arrange any 10 numbers between 1 to 80 in any order.

The number of combinations can be found using the following formula:

[number of options] ^ [number of digits]

80 ^ 10

So your answer is:

10,737,418,240,000,000,000

Each pair of digits has 170 posssible permutations of the numbers 1 thru 80, so the total number of permutations is 170^5 = 141,985,700,000

10^80=1.e+80 :))) you need alot of paper to count!