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Thinking of the problem as a nested for loop it's intuitive to see how you get 1000 combinations without the numbers being unique but I dont see how to get 720

Thinking of the problem as a nested for loop it's intuitive to see how you get 1000 combinations without the numbers being unique but I dont see how to get 720 combinations as the answer when each digit is different. Can you explain where 720 combinations comes from in the context I mentioned?

mooreds 8y ago
10 * 9 * 8 The first time you pick, you have 10 different numbers to choose from, the second time, 9, the third time 8.
- somehnreader 8y ago
  Or if you are like me (and a bit stupid): 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.
  - mooreds 8y ago
    Hmmm. I wouldn't have approached it that way. Thanks for sharing! Lots of ways to approach any problem.
deleted 8y ago
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the_jeremy 8y ago
for i = 1-10 for j = 1-10 if i == j break for k = 1-10 if i == j or i == k break log(i,j,k)
the_jeremy 8y ago
for i = 1-10 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)
- m463 8y ago
  #!/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))
joshvm 8y ago
The questions ask for all permutations of length 3. What you've described is the Cartesian product: 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 itertools.product essentially does a nested for loop. The reason it's less is because each number has to be different, so you can't have 000 or 111, etc.
eam 8y ago
Permutation without repetition. 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.
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ggggtez 8y ago
I'd like to highlight this comment for everyone who is saying the question can only be gotten wrong by someone who misreads it. It can be easy for people to forget that many people (even coders) have never done this type of math before.
- cdf2theworld 8y ago
  I believe CS degrees require discreet math (and highschool algebra 2), which means they've done permutations and combinatorics, so yes they've done these types of problems before. The real question is how long has it been?

mooreds 8y ago

10 * 9 * 8 The first time you pick, you have 10 different numbers to choose from, the second time, 9, the third time 8.

somehnreader 8y ago
Or if you are like me (and a bit stupid): 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.
- mooreds 8y ago
  Hmmm. I wouldn't have approached it that way. Thanks for sharing! Lots of ways to approach any problem.

deleted 8y ago

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the_jeremy 8y ago

for i = 1-10 for j = 1-10 if i == j break for k = 1-10 if i == j or i == k break log(i,j,k)

the_jeremy 8y ago

for i = 1-10 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)

m463 8y ago
#!/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))

joshvm 8y ago

The questions ask for all permutations of length 3. What you've described is the Cartesian product: 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 itertools.product essentially does a nested for loop. The reason it's less is because each number has to be different, so you can't have 000 or 111, etc.

eam 8y ago

Permutation without repetition. 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.

deleted 8y ago

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ggggtez 8y ago

I'd like to highlight this comment for everyone who is saying the question can only be gotten wrong by someone who misreads it. It can be easy for people to forget that many people (even coders) have never done this type of math before.

cdf2theworld 8y ago
I believe CS degrees require discreet math (and highschool algebra 2), which means they've done permutations and combinatorics, so yes they've done these types of problems before. The real question is how long has it been?