9 Digit Number Formation Problem JEE

 

❓ Question

Let

$$S=\{1,2,3,4,5,6,7,8,9\}$$

Let $x$ be the number of 9-digit numbers formed using digits of $S$ such that exactly one digit is repeated and it is repeated exactly twice.

Let $y$ be the number of 9-digit numbers formed using digits of $S$ such that exactly two digits are repeated and each is repeated exactly twice.

Then:

1. $56x=9y$
   
2. $9x=2y$
   
3. $21x=4y$
   
4. $45x=7y$
   

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Permutations with repetition
👉 Counting arrangements
👉 Relation between two counting cases.

Main idea:

Find $x$ and $y$ separately using multinomial counting.

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🔷 Step 1 — Find $x$ 💯

We need 9-digit numbers using digits $1$ to $9$.

Exactly one digit repeats twice.

So total digits used:

$$8 \text{ distinct digits}$$

because one digit appears twice.

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Choose repeated digit

$${}^9C_1=9$$

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Choose remaining 7 digits

From remaining 8 digits:

$${}^8C_7=8$$

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Arrange 9 digits

Arrangement of:

$$9 \text{ digits with one repeated twice}$$
$$= \frac{9!}{2!}$$

Thus:

$$x = 9\times8\times\frac{9!}{2}$$
$$\boxed{ x=36\cdot9! }$$

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🔷 Step 2 — Find $y$

Now exactly two digits are repeated twice each.

So total digits:

$$7 \text{ distinct digits}$$

because:

$$2+2+5=9$$

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Choose two repeated digits

$${}^9C_2$$

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Choose remaining 5 digits

From remaining 7 digits:

$${}^7C_5$$

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Arrange digits

Now arrangement of:

$$9 \text{ digits with two pairs repeated}$$
$$= \frac{9!}{2!2!}$$

Hence:

$$y = {}^9C_2\cdot{}^7C_5\cdot\frac{9!}{4}$$

Now:

$${}^9C_2=36$$
$${}^7C_5=21$$

Thus:

$$y = 36\times21\times\frac{9!}{4}$$
$$= 189\cdot9!$$

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🔷 Step 3 — Compare $x$ and $y$

$$x=36\cdot9!$$
$$y=189\cdot9!$$

Thus:

$$\frac{y}{x} = \frac{189}{36} = \frac{21}{4}$$

So:

$$\boxed{ 4y=21x }$$

or:

$$\boxed{ 21x=4y }$$

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✅ Final Answer

$$\boxed{ 21x=4y }$$

(Option 3)

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🔷 JEE Trap Alert 🚨

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Remember:

• One repeated pair:

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

• Two repeated pairs:

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

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