Question

Combinations- how to show that 10C2=10C8

Answer 1

By definition,


`{}^nC_k=\dbinom nk=(n!)/(k!(n-k)!)`

We have


`\dbinom{10}2=(10!)/(2!(10-2)!)=(10!)/(2!8!)=(10!)/((10-8)!8!)=\dbinom{10}8`

and so


`{}^(10)C_8={}^(10)C_2`