# A die is rolled 4 times. The probability of getting a larger number than the previous number each time is

Question : A die is rolled 4 times. The probability of getting a larger number than the previous number each time is

(A) 17/216

(B) 5/432

(C) 15/432

(D) none of these

Given that, n(S) = 6 × 6 × 6 × 6 = $6^{4}$
The number of favourable ways is $6C_{4}=\dfrac{6\times 5}{2}=15$
$\dfrac{15}{6\times 216}-\dfrac{5}{2\times 216}\Rightarrow \dfrac{5}{432}$