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Of the 9-letter passwords formed by rearranging the letters AAAABBCCC (4 A's, 2 B's, and 3 C's), I select one at random. Determine the following probabilities. (a) Prob (my word is a palindrome and has no two C's next to each other). 4 points (b) Prob (my word has two C's next to each other and the other C not next to them). (C) Prob (my word has the three C's next to each other and the B's apart from each other). Leave your probabilities with factorials in them.

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Answer:

a) 3

b) (8!/9!)-(7!/9!)

c) (1-(8!/9!))*(7!/9!)

Explanation:

a)With 4 As ; 2Bs and 3Cs it is possible to get a palindrome if you fixed the letters C according to: (2) in the extremes of the word and the other one at the center therefore you only have palindrome in the following cases

C ( ) C ( ) C

To fill in the gaps we have 4 letters A and 2 letters B, wich we have two divide in two palindrome gaps,

AAB and BAA the palindrome is C AAB C BAA C

BAA and AAB " " is C BAA C AAB C

ABA and ABA " " is C ABA C ABA C

b) 4 A ; 2B ; 3C

We have the total number of elements 9, so the total number of possible outcomes is : 9!

Total events: 9!

if we fixed 3 C we have (the group of 3 Cs becoming one element) so the total amount of events with 3 adjacent Cs is: 7!

Therefore the probability of having 3 adjacent Cs is: 7!/9!

If we fixed only 2 Cs we have:

4 A ; 2 B ; 2C : 1C

Total number of words (events) in this case is 8! (2C becomes 1 element)

so the total numbers of events is 8! the probability in this case is 8!/9!(this value includes cases of adjacent 3 Cs previous calculated ) so this value minus the case of 3 adjacent Cs ) give us 2 adjacent C and the other no next to them

Probability (of words with 2 adjacent Cs and the other no next to them is); 8!/9! - 7!/9!

c) Probability of B apart from each other is the whole set of events minus those where 2 B are adjacent or (become 1 element)

4 A ; 2B ; 3C

Total of events 9! and events with adjacent B is 8!/9!

Therefore the probability of words with 3 adjacent Cs and 2 B separeted is

the probability of 3 adjacent Cs (7!/9!) times probability of words with no adjacent Bs wich is (1-(8!/9!))*(7!/9!)

User Carl Vitullo
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