   SEARCH HOME Math Central Quandaries & Queries  Question from Alfred, a parent: A briefcase lock has 3 columns of numbers each 0-9. It opens with the correct combination of three numbers. How many combinations are possible to open the lock. Thank you. Hi Alfred,

Set the first two columns to 0 and cycle the third column through the 10 possibilities, 0, 1, 2, ... ,9. This gives you 10 possible combinations. Leave the first column at 0, set the second column to 1 and again cycle the third column through the 10 possibilities. This give 10 more possible combinations. Leave the first column at 0, set the second column to 2 then 3 then 4 and so on, each time cycling the third column through the 10 possibilities. Each time you get 10 new possible combinations so you have now tied 10 × 10 = 100 possible combinations.

Each of the combinations you have tried have the first column set to 0. Set the first column to 1 and start again to get a further 100 possible combinations. Now set the first column to 2 and then 3 and so on. Each time you get 100 possible combinations so all together you have 10 × 100 = 10 × 10 × 10 = 1,000 possible combinations.

There is another way to see this. Each of the possible combinations is a 3 digit number if you allow me to write leading zeros. By that I mean I will write five as 005 and twenty-three as 023 etc. The smallest 3 digit number I have is zero, written 000 and the largest is 999 and I have ever integer between 000 and 999. Thus I have 1,000 numbers or 1,000 possible combinations.

I hope this helps,
Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.