'Simple' permutation problem

1. Jun 24, 2010

majormuss

1. The problem statement, all variables and given/known data
In how many ways can this selection: {0,1,2,3,4,5} be written in a 3 digit form?(without repetition)

2. Relevant equations

3. The attempt at a solution
my answer was 120 but the book says it's 100, I am confused.. need help.

2. Jun 24, 2010

Staff: Mentor

Assuming that numbers such as 012 are allowed, I get 120 also. Starting from the hundreds' place, any of the six digits can be used. In the tens' place, any of five remaining digits can be used. In th ones' place any of the four remaining digits can be used. 6 * 5 * 4 = 120.

If numbers such as 012 aren't allowed, then you have only five choices for the hundreds' digit, five for the tens' digit and four for the ones' digit.

3. Jun 24, 2010

majormuss

why do you say 5 for the hundreds and five again for the tens place? isn't it supposed to be 5,4,3?? Also why will 012 not be used? it didn't say that in the question...

4. Jun 25, 2010

Staff: Mentor

We don't usually represent numbers with leading zeroes - that's all I'm saying. If the hundreds' digit can't be 0, then it must be 1, 2, 3, 4, or 5, so there are five possibilities. The tens' digit could be 0 or any one of 1, 2, 3, 4, or 5 that hasn't already been used. That's another five. The ones' digit could be any of 0, 1, 2, 3, 4, or 5 that hasn't already been used in the other two places. That's four possibilities.

If numbers such as 012 are OK, then there are 120 different possibilities.

5. Jun 25, 2010

HallsofIvy

The problem spoke of "three digit form" which implies we are talking about numbers, not just permutations of symbols. 012= 12 which is NOT "3 digits".

For the first digit on the left you can use any of the digits 1, 2, 3, 4, or 5. For the second digit, you can use any of the remaining digits- 0, 2, 3, 4, 5 if 1 was the first digit, 0, 1, 3, 4, 5, if 2 was the first digit, etc., but 5 digits in any case. For the last digit any of the remaining 4 digits can be used. Total number of possible numbers, 5*5*4= 100.