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asked Aug 13, 2017 in Algebra 1 Answers by anonymous

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1 Answer

If all the letters were different there would be 7! (=7*6*5*4*3*2*1) = 5040 passwords.

But 4 of the letters are A and the reduces the number of passwords by 24, because there are 24 ways of arranging 4 different objects. 5040/24=210. Also there are 2 Bs and there are 2 ways of arranging 2 different objects, so we need to divide 210 by 2 = 105. So there are 105 different 7-letter passwords.

Another way to do this is to look at the As and Bs only. With these you can make only 15 different 6-letter passwords:

AAAABB, AAABAB, AAABBA, AABAAB, AABABA, AABBAA, ABAAAB, ABAABA, ABABAA, ABBAAA, BAAAAB, BAAABA, BAABAA, BABAAA, BBAAAA.

There is only one C, and it can go in any one of 7 positions, so the total number of passwords=7*15=105.

answered Aug 14, 2017 by Rod Top Rated User (509,800 points)

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