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JEE Mathematics 2025 Question with Solution

Three distinct numbers are selected randomly from the set {1,2,3,,40}\{1, 2, 3, \dots, 40\}. If the probability that the selected numbers are in an increasing G.P. is mn\frac{m}{n}, where gcd(m,n)=1\gcd(m, n) = 1, then m+nm + n is equal to:

  • A

    1414

  • B

    3131

  • C

    1616

  • D

    1313

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Three distinct numbers are chosen from {1,2,3,,40}\{1,2,3,\dots,40\}.

Find: The value of m+nm+n if the probability of getting an increasing G.P. is mn\frac{m}{n}.

The solution counts the favorable selections and gets

Ntotal=(403)=9880N_{\text{total}}=\binom{40}{3}=9880

and the number of increasing G.P. triplets as

Nfavorable=28N_{\text{favorable}}=28

Therefore,

P=289880=72470P=\frac{28}{9880}=\frac{7}{2470}

So,

m=7,n=2470m=7,\quad n=2470

and hence

m+n=2477m+n=2477

Thus, the working shown in the solution gives 24772477. However, this does not match any of the listed options. The solution's lists option D as the correct option, so there is a clear discrepancy between the options and the extracted solution content.

Common mistakes

  • Counting ordered triples instead of selected sets is incorrect because the question asks for three numbers to be selected, not arranged. Use combinations (403)\binom{40}{3} for the total sample space.

  • Assuming only integer common ratios misses valid G.P. triplets such as those with ratio 32\frac{3}{2}. For integer terms, use the form (kq2,kqp,kp2)\left(kq^2,kqp,kp^2\right) with coprime integers p>q1p>q\ge 1.

  • Forgetting the condition that all three terms must lie in {1,2,,40}\{1,2,\dots,40\} leads to overcounting. Always impose the bound on the largest term, namely kp240kp^2 \le 40.

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