'Simple' permutation problem

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In summary, if numbers such as 012 are allowed, then you can have 120 different three digit combinations.
  • #1
majormuss
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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)



Homework Equations





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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  • #2
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
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...
 
  • #4
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.
 
  • #5
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.
 

1. What is a permutation problem?

A permutation problem is a type of mathematical problem that involves arranging or grouping a set of objects or numbers in a specific order.

2. What makes a permutation problem simple?

A permutation problem is considered simple when the objects or numbers being arranged are distinct and the order of arrangement is important. In other words, there are no restrictions or special conditions to consider.

3. How do I approach a 'simple' permutation problem?

The first step in approaching a 'simple' permutation problem is to identify the number of objects or numbers that need to be arranged and the order in which they should be arranged. Then, use the factorial formula (n!) to calculate the total number of possible arrangements.

4. What is the difference between a permutation and a combination?

A permutation involves arranging objects or numbers in a specific order, whereas a combination does not consider the order of arrangement. In other words, a combination is a subset of a permutation.

5. Can permutation problems be solved without using the factorial formula?

Yes, permutation problems can also be solved using other methods such as using a tree diagram or by using the permutation formula nPr = n!/(n-r)!. However, the factorial formula is the simplest and most commonly used method.

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