Let the no. be 10x + y.No. formed by the interchange of digits = 10y + xWe have y - x = 2 .........(i)y + x = 14 .........(ii)Solving (i) and (ii), we get x = 6, and y = 8∴ the no. is 68.
Let the middle digit be x. Then, 2x = 10 or x = 5.So, the number is either 253 or 352. Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit. Hence, required number = 253.
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
EXAMIANS