How many permutations are there for all 10 numbers from 0 to 9 taken 3 at a time, if the 3 numbers cannot begin with a zero, that is: 012 not allowed but 102...etc, allowed? I thank you for help.

Guest Mar 4, 2018

#1**+2 **

My guess

you cannot have 0 then 9 x 8 numbers = 72 numbers

out of 10 p 3 = 720 720 - 72 = 648 combos

ElectricPavlov
Mar 4, 2018

#2**+2 **

EP is right. There are 10P3 = 720 permutations of the 10 numbers from 0 to 9. Since each permutation begins with one of 10 digits, including zero, then there should be:720 / 10 =72 permutations that will begin with zero. Therefore, the total number of permutations excluding zero at beginning would be: 720 - 72 =** 648. **Here is a list of ALL numbers that begin with zero. You may count them and should get 72:

**{0, 1, 2} | {0, 1, 3} | {0, 1, 4} | {0, 1, 5} | {0, 1, 6} | {0, 1, 7} | {0, 1, 8} | {0, 1, 9} | {0, 2, 1} | {0, 2, 3} | {0, 2, 4} | {0, 2, 5} | {0, 2, 6} | {0, 2, 7} | {0, 2, 8} | {0, 2, 9} | {0, 3, 1} | {0, 3, 2} | {0, 3, 4} | {0, 3, 5} | {0, 3, 6} | {0, 3, 7} | {0, 3, 8} | {0, 3, 9} | {0, 4, 1} | {0, 4, 2} | {0, 4, 3} | {0, 4, 5} | {0, 4, 6} | {0, 4, 7} | {0, 4, 8} | {0, 4, 9} | {0, 5, 1} | {0, 5, 2} | {0, 5, 3} | {0, 5, 4} | {0, 5, 6} | {0, 5, 7} | {0, 5, 8} | {0, 5, 9} | {0, 6, 1} | {0, 6, 2} | {0, 6, 3} | {0, 6, 4} | {0, 6, 5} | {0, 6, 7} | {0, 6, 8} | {0, 6, 9} | {0, 7, 1} | {0, 7, 2} | {0, 7, 3} | {0, 7, 4} | {0, 7, 5} | {0, 7, 6} | {0, 7, 8} | {0, 7, 9} | {0, 8, 1} | {0, 8, 2} | {0, 8, 3} | {0, 8, 4} | {0, 8, 5} | {0, 8, 6} | {0, 8, 7} | {0, 8, 9} | {0, 9, 1} | {0, 9, 2} | {0, 9, 3} | {0, 9, 4} | {0, 9, 5} | {0, 9, 6} | {0, 9, 7} | {0, 9, 8}** | {1, 0, 2} | {1, 0, 3} | {1, 0, 4} | {1, 0, 5} | {1, 0, 6} | {1, 0, 7} | {1, 0, 8} | {1, 0, 9} | ... (total: 720)

Guest Mar 4, 2018

#3**0 **

Holy Guacamole ! I hope you didn't TYPE all of those ! I'd have at LEAST 10 C 3 typos !

ElectricPavlov
Mar 4, 2018