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Possible Permutations for 6-Position Car Plate: Numbers and Alphabets (ABC 123)
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[QUOTE="ZeroPivot, post: 4493636, member: 487109"] [h2]Homework Statement [/h2] Lets say a car plate has 6 positions, three of them are numbers and 3 of them are alphabetic. e.g. ABC 123. I need to figure out how many possible permutations can occur using all the numbers and alphabets BUT i cannot use any number or alfabet twice in an permutation eg. AAB 123 or ABC 112 is forbidden. [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] first i split the problem, first the numbers then the alfabet: NUMBERS: so N1 is the number of numbers i can use which is 10 obviously. 0-9 K is the number of positions which is 3. N1!/(N1-K)=10!/(10-3)! ALFABET: so N2 is the number of alfabets which is 26, K is the number of positions 3. N2!/(N2-K)!=26!/(26-3)! The ANSWER: (N1!/(N1-K))*(N2!/(N2-K)!) = (10!/(10-3)!)*(26!/(26-3)!) is the answer correct? [/QUOTE]
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Possible Permutations for 6-Position Car Plate: Numbers and Alphabets (ABC 123)
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