Guys, some help... Whats the best math formula to resolve this

>>56405710
>not accepting only zero nor zeros to the left
So 0, 00, 000, etc. don't count? 100, 386, count, but 0100, 00100, 00386 do not count?

>>56405935
yes, its the all the set W is all about.

https://www.cymath.com/

wut ?

>>56405953
Okay, well, you need to have at least one non-zero digit. You can pick this in 9 ways (one for each digit 1 to 9). Since you want a number less than 1000000, you can have at most 5 digits.
Thus, you can pick the first digit in 9 ways: one for each digit from 1 to 9. The rest of the digits are chosen from the remaining 9 (0 to 9, excluding the one your first picked). Order matters, so you use permutations:
You can pick k digits in 9*P (9,k-1) ways, so just sum that from k = 0 to 5

>>56405710
Why did you write 9-(x-1) and not 10-x tho

>>56405710
Your solution's right. K is the amount of numbers you can make that don't have repeating digits and W is the amount of those numbers that start with a zero, which is a tenth. So K-W is the amount of non-repeating numbers that don't have 0's on the left.

hi

File: 1470499459238.gif (2MB, 500x390px) Image search: [Google]

2MB, 500x390px

>>56406412
to keep the same ratio 1~6, 1 meaning one digit and 6 the maximum allowed digit (last number would be 999999, 6 digits).

But i guess there's no math formula other than summation of combinations. Well, thanks anyway.

>>56405710

1) This belongs to >>>/sci/

2) I smell homework

3) It's a really simple task, statistics 101:


First we get all numbers with 6 digits:

The first digit has 9 possibilities (1-9), second number also 9 (0-9 minus the previous one) and so on - so it's "9 * 9 * 8 * 7 * 6 * 5" possibilities.

And so on.


No fuck off.

>>56407354

By the way, I get the same result as you:

(9 * 9 * 8 * 7 * 6 * 5)
+ (9 * 9 * 8 * 7 * 6)
+ (9 * 9 * 8 * 7)
+ (9 * 9 * 8)
+ (9 * 9)
+ 9
=
136,080
+ 27,216
+ 4,536
+ 648
+ 81
+ 9
=
168570

>>56407354
yes man, i did it, in case you didn't notice its answered, i'm just looking to improve my answer.

i can do some simplification as:

import math
f = math.factorial
r = 0

for i in range(1,7):
    r+=(f(10)-f(9))/f(10-1)

print(r)

>>56407539
i mean:

for i in range(1,7):
    r+=(f(10)-f(9))/f(10-i)

>>56405710
why isn't the answer 999999 for all numbers below 100000 that aren't 0?

>>56407602
can't repeat digits 0~9
its just 987654 the higher one

>>56407539

You can also write it like this:

9 * 9 * 8 * 7 * 6 * 5 ( 1 + 1/5 + 1/(6*5) + 1/(7*6*5) + 1/(8*7*6*5) + 1/(9*8*7*6*5) )


Then you can transform it into this:

9 * (9!/4!) * ( 1/(4!/4!) + 1/(5!/4!) + 1/(6!/4!) + 1/(7!/4!) + 1/(8!/4!) + 1/(9!/4!) )


Which is the same as:

9 * (9!/4!) * 4! * (1/4! + 1/5! + 1/6! + 1/7! + 1/8! + 1/9!)


And finally:

9 * 9! * (1/4! + 1/5! + 1/6! + 1/7! + 1/8! + 1/9!)


I doubt it gets any shorter than this.