Answer:
a) 3
b) (8!/9!)-(7!/9!)
c) (1-(8!/9!))*(7!/9!)
Step-by-step 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
<u>C</u> ( ) <u>C</u> ( ) <u>C</u>
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!)
)²on both side