JEE Main 2023 · 29 Jan · Shift 1 · numerical answer

JEE Main 2023 Mathematics question on Permutations and Combinations

If all the six digit numbers x1x2x3x4x5x6x_1 x_2 x_3 x_4 x_5 x_6 with 0<x1<x2<x3<x4<x5<x60 < x_1 < x_2 < x_3 < x_4 < x_5 < x_6 are arranged in the increasing order, then the sum of the digits in the 72nd72^{\text{nd}} number is ________.
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Answer: 32

Solution

We need six-digit numbers with strictly increasing digits chosen from 1–9 (since x₁ > 0, all digits are from 1–9).

Total such numbers = C(9,6) = 84, each corresponding to a unique 6-element subset of {1,...,9} arranged in increasing order.

Finding the 72nd number:

Numbers starting with 1: fix x₁=1, choose 5 from {2,...,9} = C(8,5) = 56 numbers (positions 1–56).
Numbers starting with 2: fix x₁=2, choose 5 from {3,...,9} = C(7,5) = 21 numbers (positions 57–77).

So the 72nd number starts with 2. We need the (72−56) = 16th number among those starting with 2.

Numbers starting with 2,3: choose 4 from {4,...,9} = C(6,4) = 15 numbers (positions 1–15 within this group).
Numbers starting with 2,4: choose 4 from {5,...,9} = C(5,4) = 5 numbers (positions 16–20).

So the 16th number starts with 2,4. We need the 1st number among those starting with 2,4.

The 1st such number starting with 2,4 is the smallest: choose the 4 smallest from {5,6,7,8,9} = {5,6,7,8}.

So the 72nd number is 2,4,5,6,7,8 → 245678.

**Sum of digits = 2+4+5+6

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