From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\frac{\mathrm{a}}{\mathrm{b}}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{N}$ and $\operatorname{gcd}(\mathrm{a}, \mathrm{b})=1$, then $\mathrm{a}+\mathrm{b}$ is equal to ______.

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Probability

From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to

\frac{a}{b}

, where

a, b \in N

and

gcd (a, b) = 1

, then

a + b

is equal to ______.

Answer: 944

Three dates can be chosen from

31

(3 31) = 4495

ways. A set of three distinct dates is in increasing A.P. exactly when, written as

a, a + d, a + 2 d

with

d \geq 1

, the terms stay within

1

31

; for common difference

d

there are

31 - 2 d

choices of

a

, so the count is

\sum_{d = 1}^{15} (31 - 2 d) = 29 + 27 + \dots + 1 = 1 5^{2} = 225

. The probability is

\frac{225}{4495} = \frac{45}{899}

, and since

899 = 29 \cdot 31

shares no factor with

45

a = 45

b = 899

. Hence

a + b = 944

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Integer answer