How Many Ways to Form a 3-Digit Number from {0,1,2,3,4,5} Without Repetition?

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The discussion centers around calculating the number of ways to form a 3-digit number from the set {0,1,2,3,4,5} without repetition. There is a debate on whether numbers with leading zeros, such as 012, should be considered valid. If leading zeros are allowed, the total combinations amount to 120, calculated as 6 choices for the first digit, 5 for the second, and 4 for the third. However, if leading zeros are not allowed, the first digit can only be chosen from 1 to 5, resulting in 100 valid combinations when calculated as 5 choices for the first digit, 5 for the second, and 4 for the third. The key takeaway is that the interpretation of leading zeros significantly affects the total count of valid 3-digit numbers.
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Homework Statement


In how many ways can this selection: {0,1,2,3,4,5} be written in a 3 digit form?(without repetition)



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The Attempt at a Solution


my answer was 120 but the book says it's 100, I am confused.. need help.
 
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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.
 
Mark44 said:
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.
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...
 
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 anyone 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.
 
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.
 
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