`T_n = .^nc_3`
`.^(n+1)c_3 - .^nc_3 = 10`
`((n+1)!)/(3!(n-2)!) - (n!)/(3!(n-3)!) = 10`
`((n+1)(n)(n-1))/(3!) - (n(n-1)(n-2))/(3!) = 10`
`(n+1)(n)(n-1) - n(n-1)(n-2) = 60`
`n(n-1)[n+1 - n+2] = 60`
`3n(n-1) = 60`
`n^2 - n-20 = 0`
`n^2 - 5n + 4n -20 = 0`
`(n-5)(n+4) = 0`
`n=5 as, n!= -4`
option 1 is correct