JEE Main 2023 · 10 April · morning · Single correct · +4 / −1
Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that 2^N < N! is m/n, where m and n are coprime, then 4m-3n is equal to :
- (a) 12
- (b) 6
- (c) 8
- (d) 10
Answer and solution
Answer (c) 8
Solution
Maths sir · 4 steps
Check the first few sums. Once N! overtakes 2^N it stays ahead, so only the smallest sums fail.
- Test small sums. The sum N of two dice runs from 2 to 12. Compare 2^N and N! for the smallest values of N.
N 2^N N! Is 2^N < N!? 2 4 2 no 3 8 6 no 4 16 24 yes - Stays true after. Going from N to N+1, 2^N gets multiplied by 2, but N! gets multiplied by N+1, which is at least 5. So 2^N < N! holds for every N from 4 to 12.
- Use the complement. The inequality fails only for N = 2, which is the outcome (1,1), and for N = 3, which is (1,2) or (2,1). That is 3 outcomes out of 36, so the probability is 1 - 3/36 = 33/36 = 11/12.
- So. m = 11 and n = 12, so 4m - 3n = 44 - 36 = 8. The answer is (c).
The idea to keep. When a condition fails only for a few small cases, count those failures and subtract from 1. Also reduce the fraction fully before reading off m and n.
Answer from the NTA final answer key. Solution by PrepWiser.