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Nine-digit numbers and the ratio n1/n2 for the digits at two places | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Let n=x1x2x3…x9n = x_1x_2x_3\ldots x_9 be formed using the digits 1,2,…,91,2,\ldots,9, each digit being used once only. If n1n_1 is the number of possible nn for which 1<∣x3−x4∣<51<\left|x_3-x_4\right|<5 and n2n_2 is the number of possible nn for which ∣x3−x4∣=1\left|x_3-x_4\right| = 1, then [n1n2]\left[\dfrac{n_1}{n_2}\right] is equal to (where [ ⋅ ][\,\cdot\,] is G.I.F.
Solution
Answer: 2
Step 1: Only x3x_3 and x4x_4 are constrained. Once an ordered pair of distinct digits is placed there, the remaining seven digits fill the remaining seven positions in 7!7! ways, the same factor in both counts, so it cancels in the ratio. Step 2: For n2n_2, ∣x3−x4∣=1\left|x_3-x_4\right| = 1. The unordered pairs of consecutive digits are {1,2},{2,3},…,{8,9}\{1,2\},\{2,3\},\ldots,\{8,9\}, eight of them, and each can be placed in 22 orders:
n2=2×8×7!=16⋅7!n_2 = 2\times8\times7! = 16\cdot7!
 Step 3: For n1n_1, the inequality 1<∣x3−x4∣<51<\left|x_3-x_4\right|<5 is strict at both ends, so the difference is 22, 33 or 44 and nothing else. For a difference dd the unordered pairs are {k,k+d}\{k,k+d\} with 1≤k≤9−d1 \le k \le 9-d, so there are 9−d9-d of them:
d=2: 7,d=3: 6,d=4: 5d=2:\ 7 ,\qquad d=3:\ 6 ,\qquad d=4:\ 5
 Doubling for order,
n1=2(7+6+5)×7!=36⋅7!n_1 = 2\left(7+6+5\right)\times7! = 36\cdot7!
 Step 4: The 7!7! cancels.
n1n2=3616=94=2.25\frac{n_1}{n_2} = \frac{36}{16} = \frac94 = 2.25
[2.25]=2\left[2.25\right] = 2
 (Running over all 9×8=729\times8 = 72 ordered pairs (x3,x4)\left(x_3,x_4\right) of distinct digits gives 3636 and 1616 directly.) Answer: 22
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