Question

Show that
C₀ + C₁ + C₂ + ... + C₉ = 512

Answer 1

hope this helps you..nC0+nC1.....nCn is generally written in the form...C0+C1.....Cn

Answer 2

Answer:

⁹C₀ + ⁹C₁   + ⁹C₂  + ⁹C₃ + ⁹C₄ + ⁹C₅ + ⁹C₆ + ⁹C₇ + ⁹C₈ + ⁹C₉   = 512

Step-by-step explanation:

⁹C₀ + ⁹C₁   + ⁹C₂  + ⁹C₃ + ⁹C₄ + ⁹C₅ + ⁹C₆ + ⁹C₇ + ⁹C₈ + ⁹C₉   = 512

ⁿCₐ = n!/(a!(n-a)!)

LHS

= ⁹C₀ + ⁹C₁   + ⁹C₂  + ⁹C₃ + ⁹C₄ + ⁹C₅ + ⁹C₆ + ⁹C₇ + ⁹C₈ + ⁹C₉

= 9!/(0!(9-0)!) + 9!/(1!(9-1)!)  + 9!/(2!(9-2)!)  + 9!/(3!(9-3)!)  + 9!/(4!(9-4)!) + 9!/(5!(9-5)!)  + 9!/(6!(9-6)!)  + 9!/(7!(9-7)!)  + 9!/(8!(9-8)!) + 9!/(9!(9-9)!)

= 9!/(0!(9)!) + 9!/(1!(8)!)  + 9!/(2!(7)!)  + 9!/(3!(6)!)  + 9!/(4!(5)!) + 9!/(5!(4)!)  + 9!/(6!(3)!)  + 9!/(7!(2)!)  + 9!/(8!(1)!) + 9!/(9!(0)!)

= 1  + 9  + 9*8/2   + 9*8*7/(3*2)  + 9*8*7*6/(4*3*2)  + 9*8*7*6/(4*3*2) + 9*8*7/(3*2) + 9*8/2 + 9 + 1

= 1 + 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1

= 512

= RHS

Hence

⁹C₀ + ⁹C₁   + ⁹C₂  + ⁹C₃ + ⁹C₄ + ⁹C₅ + ⁹C₆ + ⁹C₇ + ⁹C₈ + ⁹C₉   = 512

Also we can say that :

⁹C₀ + ⁹C₁   + ⁹C₂  + ⁹C₃ + ⁹C₄ + ⁹C₅ + ⁹C₆ + ⁹C₇ + ⁹C₈ + ⁹C₉is expansion of ( x + y) ⁹  where x & y = 1

= ( 1 + 1)⁹

= 2⁹

= 512