For a number to be divisible by 4, the number formed by the last two digits of that number must be divisible by 4.For a number to be divisible by 9, the sum of its digits must be divisible by 9.Among the choices, choices (a) and (b) are divisible by 4 but of the first two choices, only choice (b) is divisible by 9.Hence choice (b) is divisible by both 4 and 9.
Total balls = 5 + 4 + 4 = 13∴ n(S) = 13C3 = 2862 blue balls can be selected from 4 balls in 4C2 = 6 ways and the remaining one ball is to be selected from 9 balls in 9C1 = 9 ways ∴ n(E) = 6 x 9∴ P(E) = 6 x 9/286 = 27/143
Let the number be 10x + y.Therefore, x + y = 9 ..................(i)10x + y - (10y + x) = 4510x + y - 10y - x = 45Therefore, x - y = 5 ..................(ii)From (i) and (ii), x = 7, y = 2Number = 72Therefore, = x x y = 7 x 2 = 14
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