Question

1. Find the number of distinguishable permutations of the given letters AAABBBCD''.

Answer 1

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

= $\frac{8!}{3!3!}$

=1120

2.If  it begins with at least 2 A's , no. of permutations

$\frac{6!}{3!}$

=120

Therefore, the probability of random permutations of AA

=120/1120= 3/28= 0.1071