Three distinct numbers are selected randomly from the set \( \{1, 2, 3, \dots, 40\} \). If the probability, that the selected numbers are in an increasing G.P. is \( \frac{m}{n} \), where \( \gcd(m, n) = 1 \), then \( m + n \) is equal to:
Three distinct numbers are selected randomly from the set \( \{1, 2, 3, \dots, 40\} \). If the probability, that the selected numbers are in an increasing G.P. is \( \frac{m}{n} \), where \( \gcd(m, n) = 1 \), then \( m + n \) is equal to:
Show Hint
Correct Answer: 2477
Solution and Explanation
We need to find the probability that three distinct numbers selected randomly from the set \( \{1, 2, 3, ..., 40\} \) form an increasing Geometric Progression (G.P.). This probability is given as \( \frac{m}{n} \) in simplest form, and we must find the value of \( m + n \).
Concept Used:
The solution uses the following concepts:
1. Combinations: The total number of ways to select \(k\) items from a set of \(n\) distinct items is given by \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
2. Geometric Progression (G.P.): Three distinct numbers \(a, b, c\) are in an increasing G.P. if \( b^2 = ac \) and \( a < b < c \). The common ratio \( r = \frac{b}{a} = \frac{c}{b} \) must be greater than 1. Since \(a, b, c\) are integers, the ratio \(r\) must be a rational number, say \( r = \frac{p}{q} \) where \( p, q \) are coprime integers with \( p > q \ge 1 \). The terms of the G.P. are \( a, a\frac{p}{q}, a\frac{p^2}{q^2} \). For all terms to be integers, \( q^2 \) must divide \( a \). Thus, we can write \( a = k \cdot q^2 \) for some integer \( k \ge 1 \), which makes the triplet \( (kq^2, kqp, kp^2) \).
Step-by-Step Solution:
Step 1: Calculate the total number of ways to select three distinct numbers.
The total number of outcomes is the number of ways to choose 3 numbers from the set of 40 numbers, which is:
\[ N_{\text{total}} = \binom{40}{3} = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 40 \times 13 \times 19 = 9880 \]
Step 2: Find the number of favorable outcomes by enumerating all possible increasing G.P. triplets.
We need to find triplets \( (a, b, c) \) from the set \( \{1, 2, ..., 40\} \) such that \( a < b < c \) and \( b^2 = ac \). We can systematically find these by considering the common ratio \( r = p/q \). The triplet form is \( (kq^2, kqp, kp^2) \) and the condition is that the largest term, \( kp^2 \), must be less than or equal to 40.
Step 3: Enumerate triplets with an integer common ratio (\( q=1 \)).
The triplet is of the form \( (k, kr, kr^2) \) with \( kr^2 \le 40 \).
For a ratio of \( r=2 \), we need \( 4k \le 40 \implies k \le 10 \). This gives 10 triplets.
For a ratio of \( r=3 \), we need \( 9k \le 40 \implies k \le 4 \). This gives 4 triplets.
For a ratio of \( r=4 \), we need \( 16k \le 40 \implies k \le 2 \). This gives 2 triplets.
For a ratio of \( r=5 \), we need \( 25k \le 40 \implies k \le 1 \). This gives 1 triplet.
For a ratio of \( r=6 \), we need \( 36k \le 40 \implies k \le 1 \). This gives 1 triplet.
Total for integer ratios = \( 10 + 4 + 2 + 1 + 1 = 18 \).
Step 4: Enumerate triplets with a non-integer rational common ratio (\( q>1 \)).
The triplet is of the form \( (kq^2, kqp, kp^2) \) with \( kp^2 \le 40 \).
For a ratio of \( r = \frac{3}{2} \), the triplet is \( (4k, 6k, 9k) \). We need \( 9k \le 40 \implies k \le 4 \). This gives 4 triplets.
For a ratio of \( r = \frac{4}{3} \), the triplet is \( (9k, 12k, 16k) \). We need \( 16k \le 40 \implies k \le 2 \). This gives 2 triplets.
For a ratio of \( r = \frac{5}{2} \), the triplet is \( (4k, 10k, 25k) \). We need \( 25k \le 40 \implies k \le 1 \). This gives 1 triplet.
For a ratio of \( r = \frac{5}{3} \), the triplet is \( (9k, 15k, 25k) \). We need \( 25k \le 40 \implies k \le 1 \). This gives 1 triplet.
For a ratio of \( r = \frac{5}{4} \), the triplet is \( (16k, 20k, 25k) \). We need \( 25k \le 40 \implies k \le 1 \). This gives 1 triplet.
For a ratio of \( r = \frac{6}{5} \), the triplet is \( (25k, 30k, 36k) \). We need \( 36k \le 40 \implies k \le 1 \). This gives 1 triplet.
Total for non-integer ratios = \( 4 + 2 + 1 + 1 + 1 + 1 = 10 \).
Step 5: Calculate the total number of favorable outcomes and the probability.
The total number of favorable outcomes is the sum from both cases:
\[ N_{\text{favorable}} = 18 + 10 = 28 \]
The required probability is:
\[ P = \frac{N_{\text{favorable}}}{N_{\text{total}}} = \frac{28}{9880} \]
Simplifying the fraction:
\[ \frac{28}{9880} = \frac{7}{2470} \]
We are given that \( \gcd(m, n) = 1 \). Since 7 is a prime number and 2470 is not divisible by 7 (as \( 2470 = 7 \times 352 + 6 \)), the fraction is in its simplest form. So, \( m = 7 \) and \( n = 2470 \).
Final Computation & Result:
We are asked to find the value of \( m + n \).
\[ m + n = 7 + 2470 = 2477 \]
The value of \( m + n \) is 2477.
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