If `.^(15)C_(3r)=.^(15)C_(r+3),` find `.^(r)C_(2)`.

Explanation

We have, `.^(15)C_(3r)=.^(15)C_(r+3)`
`implies3r=r+3`
or `3r+r+3=15`
`implies2r=3` or 4r=12
`impliesr=(3)/(2)` or r=3
but `r in W`, so that `r ne(3)/(2)`
`thereforer=3`
then, `.^(r)C_(2)=.^(3)C_(2)=.^(3)C_(1)=3`