+0  
 
0
75
5
avatar

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
Sort: 

5+0 Answers

 #1
avatar+12266 
+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
avatar
+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
avatar+12266 
0

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

ElectricPavlov  Mar 4, 2018
 #4
avatar
0

No EP!!. The computer did!!.

Guest Mar 4, 2018
 #5
avatar+86536 
+1

We have 9   ways to fill the first position  (1-9), 9 ways to fill the second position (every number 0-9 except the first one we chose) and 8 ways to fill the third position (every number 0-9 except the first two we chose)  =  9 * 9 * 8    =  648

 

 

cool cool cool

CPhill  Mar 4, 2018

24 Online Users

avatar
New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy