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Question: How many different arrangements can be made out of the letters in the expansion A<sup>2</sup>B<sup>3...

How many different arrangements can be made out of the letters in the expansion A2B3C4, when written in full?

A

9!2!3!4!\frac{9!}{2!3!4!}

B

2! + 3! + 4! (2! + 3! + 4!)

C

2! + 3! – 4

D

9!2!+3!+4!\frac{9!}{2! + 3! + 4!}

Answer

9!2!+3!+4!\frac{9!}{2! + 3! + 4!}

Explanation

Solution

‘A2 B3 C4’ A®2 times, B®3 times, C®4 times

Total arrangement = (2+3+4)!2!3!4!\frac{(2 + 3 + 4)!}{2!3!4!}=9!2!3!4!\frac{9!}{2!3!4!}