Answer:
1. 1120
2, 0.1071
Step-by-step explanation:
1.the number of distinguishable permutations of the given letters
AAABBBCD A is repeated 3, B is also repeated thrice
= [tex]\frac{8!}{3!3!}[/tex]
=1120
2.If it begins with at least 2 A's , no. of permutations
[tex]\frac{6!}{3!}[/tex]
=120
Therefore, the probability of random permutations of AA
=120/1120= 3/28= 0.1071
Using the permutation principle, the number of distinguishable permutations and the probability that the permutation begins with atleast 2A's are :
Given the Letters ; AAABBBCD
Repeated alphabets :
A = 3 times ; B = 3 times
The number of distinguishable permutations can be defined thus :
Number of alphabets ÷ (number of A's)!(number of B's)!
8! ÷ (3!3!)
40320 ÷ 36 = 1120 permutations.
B.) Probability of randomly choosing a permutation which begins with atleast 2A's :
Number of permutations beginning with atleast 2A's :
(Number of alphabets - 2)! ÷ (number of A's)!
(8 - 2)! ÷ 3!
6! ÷ 3!
720 ÷ 6 = 120
P(begins with Atleast 2A's) = (number of permutation beginning with atleast 2A's) ÷ (total number of permutations)
Therefore, P(begins with Atleast 2A's) = (120) ÷ 1120 = 0.107
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