Try to clear this factorial mess!

Number Theory Level 2

Simplify and give your answer as an integer:

( 10 ! + 9 ! ) ( 8 ! + 7 ! ) ( 6 ! + 5 ! ) ( 4 ! + 3 ! ) ( 2 ! + 1 ! ) ( 10 ! − 9 ! ) ( 8 ! − 7 ! ) ( 6 ! − 5 ! ) ( 4 ! − 3 ! ) ( 2 ! − 1 ! ) . \large \dfrac{\color{#3D99F6}(10! + 9!)(8! + 7!)(6! + 5!)(4! + 3!)(2! + 1!)}{\color{#D61F06}(10! - 9!)(8! - 7!)(6! - 5!)(4! - 3!)(2! - 1!)}.


The answer is 11.

Syed Hamza Khalid by Syed Hamza Khalid , 17, France

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1 solution

Hana Wehbi
Nov 11, 2017

( 10 ! + 9 ! ) ( 8 ! + 7 ! ) ( 6 ! + 5 ! ) ( 4 ! + 3 ! ) ( 2 ! + 1 ! ) ( 10 ! − 9 ! ) ( 8 ! − 7 ! ) ( 6 ! − 5 ! ) ( 4 ! − 3 ! ) ( 2 ! − 1 ! ) = \large \dfrac{\color{#3D99F6}(10! + 9!)(8! + 7!)(6! + 5!)(4! + 3!)(2! + 1!)}{\color{#D61F06}(10! - 9!)(8! - 7!)(6! - 5!)(4! - 3!)(2! - 1!)} =

9 ! ( 10 + 1 ) 7 ! ( 8 + 1 ) 5 ! ( 6 + 1 ) 3 ! ( 4 + 1 ) 1 ! ( 2 + 1 ) 9 ! ( 10 − 1 ) 7 ! ( 8 − 1 ) 5 ! ( 6 − 1 ) 3 ! ( 4 − 1 ) 1 ! ( 2 − 1 ) = \large \dfrac{\color{#3D99F6}9!(10 + 1)7! (8+ 1)5!(6+1)3!(4+1)1!(2+1)}{\color{#D61F06}9!(10-1)7!(8-1)5!(6-1)3!(4-1)1!(2-1)}=

After cancelling the factorial terms from the Numerator and Denominator we get:

11 × 9 × 7 × 5 × 3 9 × 7 × 5 × 3 × 1 = 11 \frac{ 11\times\color{#D61F06} 9\times 7\times 5\times 3}{\color{#D61F06}9\times 7\times 5\times3\times1}= 11

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