Total balls = 4 + 6 + 7 = 17∴ n(S) = 17C1 = 680Two red balls can be selected from four red balls in 4C2 = 6 ways.and the third ball can be selected from the remaining 13 balls in 13C1 = 13 ways.∴ P (E) = 13x6/680 = 39/340
Let the digits be x and yTherefore, x + y = 12 .............(1)(10y + x) - (10x + y) Therefore, y - x = 4............. (2)Solving (1) and (2), y = 8 Therefore, x = 4There are two possible numbers 48 and 84. So the lowest no. is 48.