Of the 9-letter passwords formed by rearranging the letters bbccc (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.

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

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!)

i'm going to search it

step-by-step explanation:

step-by-step explanation: