JEE Main 2026 — Jan 28, Shift 1 — Mathematics solution

JEE Main2026Jan 28, Shift 1Mathematics

Let $S = {1, 2, 3, 4, 5, 6, 7, 8, 9}$ . Let $x$ be the number of 9-digit numbers formed using the digits of the set $S$ such that only one digit is repeated and it is repeated exactly twice. Let $y$ be the number of 9-digit numbers formed using the digits of the set $S$ such that only two digits are repeated and each of these is repeated exactly twice. Then,

A
$29 x = 5 y$
B
$45 x = 7 y$
C
$21 x = 4 y$
D
$56 x = 9 y$

Solution

For $x$ : choose the digit to repeat ( $(1 9)$ ), choose 7 more distinct digits from the remaining 8 ( $(7 8)$ ), then arrange 9 symbols with one pair: $\frac{9 !}{2 !}$ .

$x = (1 9) (7 8) \cdot \frac{9 !}{2 !} = 9 \cdot 8 \cdot \frac{9 !}{2}$ .

For $y$ : choose the 2 digits that get repeated ( $(2 9)$ ), choose 5 more distinct from the remaining 7 ( $(5 7)$ ), then arrange 9 symbols with two pairs: $\frac{9 !}{2 ! \cdot 2 !}$ .

$y = (2 9) (5 7) \cdot \frac{9 !}{4}$ .

Take the ratio:

$\frac{x}{y} = \frac{9 \cdot 8 \cdot 9 ! /2}{36 \cdot 21 \cdot 9 ! /4} = \frac{9 \cdot 8 \cdot 2}{36 \cdot 21} = \frac{144}{756} = \frac{4}{21}$ .

$∴ 21 x = 4 y$ . Option (3).

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