Factorials

Probability Level 3

The sum 1 1 ! 9 ! + 1 3 ! 7 ! + 1 5 ! 5 ! + 1 7 ! 3 ! + 1 1 ! 9 ! \large{\frac{1}{1!9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{7!3!}+\frac{1}{1!9!}} can be written in the form of 2 a b ! \large\frac{2^a}{b!} , where a a and b b are positive integers.

Find a + b a+b


The answer is 19.

Harsh Shrivastava by Harsh Shrivastava , 17, India

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1 solution

X X
Jul 1, 2018

1 1 ! 9 ! + 1 3 ! 7 ! + 1 5 ! 5 ! + 1 7 ! 3 ! + 1 1 ! 9 ! = 10 ! 1 ! 9 ! + 10 ! 3 ! 7 ! + 10 ! 5 ! 5 ! + 10 ! 7 ! 3 ! + 10 ! 1 ! 9 ! 10 ! = ( 10 1 ) + ( 10 3 ) + ( 10 5 ) + ( 10 7 ) + ( 10 9 ) 10 ! = 2 9 10 ! \dfrac{1}{1!9!}+\dfrac{1}{3!7!}+\dfrac{1}{5!5!}+\dfrac{1}{7!3!}+\dfrac{1}{1!9!}=\dfrac{\dfrac{10!}{1!9!}+\dfrac{10!}{3!7!}+\dfrac{10!}{5!5!}+\dfrac{10!}{7!3!}+\dfrac{10!}{1!9!}}{10!}=\dfrac{\binom{10}{1}+\binom{10}{3}+\binom{10}{5}+\binom{10}{7}+\binom{10}{9}}{10!}=\dfrac{2^9}{10!}

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