JEE Mathematics 2026 Question with Solution

Given: $S=\{1,2,3,4,5,6,7,8,9\}$. We define $x$ as the number of $9$-digit numbers in which exactly one digit is repeated twice, and $y$ as the number of $9$-digit numbers in which exactly two digits are repeated twice.

Find: The correct relation between $x$ and $y$.

Concept: This is a permutation problem with repeated digits. Distinct arrangements are counted by dividing by factorials of identical repeated digits.

For $x$:

• Choose the digit that is repeated twice: $\binom{9}{1}$
• Choose the remaining $7$ distinct digits from the remaining $8$ digits: $\binom{8}{7}$
• Arrange these $9$ digits, where one digit is repeated twice:

$$x=\binom{9}{1}\binom{8}{7}\frac{9!}{2!}$$

For $y$:

• Choose the two digits that are repeated twice: $\binom{9}{2}$
• Choose the remaining $5$ distinct digits from the remaining $7$ digits: $\binom{7}{5}$
• Arrange these $9$ digits, where two digits are each repeated twice:

$$y=\binom{9}{2}\binom{7}{5}\frac{9!}{2!2!}$$

Now compare $x$ and $y$:

$$\frac{x}{y}=\frac{\binom{9}{1}\binom{8}{7}\cdot 2!}{\binom{9}{2}\binom{7}{5}}=\frac{4}{21}$$

Therefore,

$$21x=4y$$

So, the correct option is A.

Given: The digits available are $1$ to $9$, with no zero involved, so every arrangement of selected digits gives a valid $9$-digit number.

Find: A relation between $x$ and $y$.

For $x$, the digit multiset has pattern $2,1,1,1,1,1,1,1$.

1. Select the repeated digit in $9$ ways.
2. Select the other $7$ digits in $\binom{8}{7}=8$ ways.
3. Arrange the resulting $9$ symbols:

$$\frac{9!}{2!}$$

Hence,

$$x=9\cdot 8\cdot \frac{9!}{2!}$$

For $y$, the digit multiset has pattern $2,2,1,1,1,1,1$.

1. Select the two repeated digits in $\binom{9}{2}$ ways.
2. Select the other $5$ digits in $\binom{7}{5}$ ways.
3. Arrange the resulting $9$ symbols:

$$\frac{9!}{2!2!}$$

Hence,

$$y=\binom{9}{2}\binom{7}{5}\frac{9!}{2!2!}$$

Now,

$$\frac{x}{y}=\frac{9\cdot 8\cdot \frac{9!}{2!}}{\binom{9}{2}\binom{7}{5}\frac{9!}{2!2!}}$$

Cancel $9!$ and simplify:

$$\frac{x}{y}=\frac{9\cdot 8\cdot 2!}{\binom{9}{2}\binom{7}{5}}=\frac{4}{21}$$

Thus,

$$21x=4y$$

Hence the correct option is A.