Total number of different combination

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The discussion focuses on calculating permutations for numbers with repeated digits. For unique digits like 4573, the total permutations are 24, calculated as 4!. When dealing with repeated digits, such as in 1966, the total permutations are 12, derived by dividing 4! by 2! to account for the overcounting of the two sixes. The conversation emphasizes the importance of adjusting calculations when multiple digits are identical, noting that different scenarios arise if three or more digits are the same. Understanding these principles is crucial for accurate permutation calculations.
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I know that if you have 4 numbers eg. 1,23,4 then the total number fo diffferent combination is
4*3*2*1 = 24
what about if one number is the same eg. 1966

Thank you
 
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Are you asking about "permutations" or "combinations"?
 
so if I had the number 4573 what would be the total permutations?
and similarly 1966 total permutations
 
msa969 said:
so if I had the number 4573 what would be the total permutations?
24, as you calculated.

msa969 said:
and similarly 1966 total permutations
12. Imagine that the two sixes are different (e.g., different colors). You start by counting 4x3x2x1, but that means that you have overcounted because 6169 is the same as 6169. You then simply divide by two to remove the overcounting: 4! / 2 = 12.
 
colours makes perfect sense
 
I believe you need to divide by 2 factorial, NOT 2.
Yes it is the same in this problem
But not if 3 or more are the same.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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