Question

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 $a/b$, where $a, b \in \mathbb{N}$ and $\gcd(a, b) = 1$, then $a + b$ is equal to _______

Answer 1

Correct Answer: 944

To solve this problem, we need to determine the probability that three randomly selected dates from a 31-day month are in an increasing arithmetic progression (A.P.). Let's break down the problem step by step:  

Step 1: Calculate Total Possible Ways to Select 3 Dates
The total number of ways to select 3 different dates from 31 is given by \( \binom{31}{3} \). This is calculated as:
\(\binom{31}{3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 4495\).

Step 2: Calculate Favorable Outcomes for Dates in Arithmetic Progression
Consider three dates \(d_1, d_2, d_3\) in arithmetic progression with a common difference \(d\), where \(d > 0\). The conditions for them to be in increasing A.P. are \(d_2 = d_1 + d\), and \(d_3 = d_1 + 2d\).
1. \(d_1\) can range from 1 to 31 - 2d as \(d_3\) must be ≤ 31.
2. \(d\) can range from 1 to 15 (since \(d = 1\) gives maximum of 29 days shift, allowing \(d_1\) to end at 29).

Calculate total favorable outcomes:
- For \(d = 1\), \(d_1\) ranges from 1 to 29 (29 values).
- For \(d = 2\), \(d_1\) ranges from 1 to 28 (28 values).
- Continue this until for \(d = 15\), \(d_1\) ranges from 1 to 1 (1 value).
Total favorable outcomes = \(29 + 28 + \dots + 1 = \frac{15 \cdot (15 + 1)}{2} = 120\).

Step 3: Calculate Probability
Probability\( = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{120}{4495}\).

Step 4: Simplify the Probability
Simplify \(\frac{120}{4495}\). Use the Euclidean algorithm to find that \(\gcd(120, 4495) = 1\), so it’s already in simplest form. The simplified probability is \(\frac{120}{4495}\).

Step 5: Find \(a + b\)
Given \(a = 120\) and \(b = 4495\), we need \(a + b = 120 + 4495 = 4615\).

Conclusion: The value of \(a + b\) is 4615, which fits within the specified range of 944 to 944.