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Compute \(\large \dfrac{(10!\color{green}{+}9!)(8!\color{green}{+}7!)(6!\color{green}{+}5!)(4!\color{green}{+}3!)(2!\color{green}{+}1!)}{(10!\color{red}{-}9!)(8!\color{red}{-}7!)(6!\color{red}{-}5!)(4!\color{red}{-}3!)(2!\color{red}{-}1!)} = \, \, \large ?\)

 Jun 29, 2020
 #1
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Numerator:  10! + 9!  =  10·9! + 9!  =  10·9! + 1·9!  =  (10 + 1)·9!  =  11·9!

                      8! + 7!  =  8·7! + 7!  =  8·7! + 1·7!  =  (8 + 1)·7!  =  9·7! 

                      6! + 5!  =  6·5! + 5!  =  6·5! + 1·5!  =  (6 + 1)·5!  =  7·5! 

                      4! + 3!  =  4·3! + 3!  =  4·3! + 1·3!  =  (4 + 1)·3!  =  5·3! 

                      2! + 1!  =  2·1! + 1!  =  2·1! + 1·1!  =  (2 + 1)·1!  =  3·1! 

 

So, the numerator becomes:  11·9! · 9·7! · 7·5! · 5·3! · 3·1!

 

Denominator:  10! - 9!  =  10·9! - 9!  =  10·9! - 1·9!  =  (10 - 1)·9!  =  9·9!

Similarly:           8! - 7!  =  7·7!

                          6! - 5!  =  5·5!

                          4! - 3!  =  3·3!

                          2! - 1!  =  1·1!

 

The denominator becomes:  9·9! · 7·7!  · 5·5! · 3·3! · 1·1!

 

Start cancelling out, and you'll get the (rather surprising) short answer.

 
 Jun 30, 2020

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