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How do I set this problem up? 1) How many even 3-digit positive integers can be written using the digits1, 2, 4, 7 & 8 ?
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are duplicates allowed?
eg 224 , 444 ?
if not then proceed as follows: the numbers must end in 2 , 4 or 8 for each of these numbers we have a number of permutations of 2 from the other 4 numbers - that is 4P2 = 12
can u continue ?
why did u set it equal to 12?
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the number of permutaions of 2 from 4 = 4! / 2! = 24 /2 = 12
and has there are 3 even nunbers the number of possibilities = 3 * 12 - this is true if no duplicates or triplicates are allowed
im confused.. the answer is 75 so im not sure how to get it
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