It can be easy for people to forget that many people (even coders) have never done this type of math before.
The first time you pick, you have 10 different numbers to choose from, the second time, 9, the third time 8.
1000 = all possible combinations from 000 to 999.
less 10 combinations with all numbers identical is 990.
now there are three patterns for numbers to be identical left that need to be taken out: 00x 0x0 x00. The x can be replaced with all digits != 0 in this example, so thats 3 (patterns) x 9 ('x' digits) = 27. That times 10 digits that can have double patterns is 270.
1000 - 10 - 270 = 720.
import itertools
print(len([x for x in itertools.product(range(10), repeat=3)])) #1000
print(len([x for x in itertools.permutations(range(10), 3)])) #720
The reason it's less is because each number has to be different, so you can't have 000 or 111, etc.
The formula is:
n! / (n − r)!
Since there is 10 (n=10) total numbers, 0-9. And it can only be 3 (r=3) digits long. Then it would be,
10! / (10-3)!* which would result in,
10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / (7 * 6 * 5 * 4 * 3 * 2 * 1) = 10 * 9 * 8 = 720.
for j = 1-10
if i == j break
for k = 1-10
if i == j or i == k break
log(i,j,k)
(I don't understand HN formatting still)
#!/usr/bin/python
r = [(i,j,k) for i in range(10) for j in range(10) for k in range(10) if i!=j and i!=k and j!=k]
print(len(r))