Concept:nCr = \(\rm {n!\over {r!\ (n - r)!}}\)Calculation:Given: nC6 : n - 3C3 = 7 : 4? \(\rm {n!\ \over 6!\ (n\ -\ 6)!} : {(n\ -\ 3)!\over 3!\ (n\ -\ 3\ -\ 3)!} = 7:4\)? \(\rm {n(n\ -\ 1)(n\ -\ 2)(n\ -\ 3)! \over 6\ \times\ 5\ \times\ 4\ \times\ 3!\ (n\ -\ 6)!} : {(n\ -\ 3)!\over 3!\ (n\ -\ 6)!} = 7:4\)? \(\rm {n(n\ -\ 1)(n\ -\ 2)\over 6\ \times\ 5\ \times\ 4} = {7\over 4}\)? n(n - 1)(n - 2) = 210n = 7 satisfy the above equation So. nC3 = 7C3 = \(\rm {7\ \times\ 6 \times\ 5 \over 3\ \times\ 2\ \times\ 1} = 35\)