Big Bang Theory cast wishes to find out the number of ways in which the word ASTRONAUT can be scrambled. They find that the number of ways in which it can be put in an unscrambling puzzle is?

Question

E and F are two independent events such that \(P(\overline{E}) = 0.6\) and \(P(E \cup F) = 0.6\). Find \(P(F)\) and \(P(\overline{E} \cup \overline{F})\).

Answer 1

Step 1: ASTRONAUT has 9 letters, with A repeated twice and T repeated twice.
Step 2: Distinct arrangements \(=\frac{9!}{2!\times2!}=\frac{362880}{4}\).
Step 3: This gives 90720.
\[ \boxed{90720} \]

Big Bang Theory cast wishes to find out the number of ways in which the word ASTRONAUT can be scrambled. They find that the number of ways in which it can be put in an unscrambling puzzle is?

The Correct Option is D

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