Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let x be the number of 9-digit numbers formed using the digits

Q: Why do we use $\binom{8}{7}$ for $x$? Shouldn't it just be $\binom{9}{7}$?

Great question. We use $\binom{8}{7}$ because one digit is already 'reserved' as the repeating one. You have 8 digits left to choose from. If you used $\binom{9}{7}$, you would be accidentally including the digit you already chose to repeat, which would violate the condition that only one digit repeats.

Q: Why do we divide by $2!2!$ for $y$?

Because the two pairs are identical within themselves. Swapping the two instances of the first pair doesn't create a new number. Since we have two such pairs, we divide by $2!$ for the first pair and $2!$ for the second pair to remove the overcounted permutations.

Q: Is there a way to solve this without factorials?

Technically, you could try to list them, but that would take a lifetime! Combinatorics is the language of factorials. Embracing them allows us to handle large numbers with elegance. Stick to the selection-arrangement framework; it is the most robust method for JEE Advanced.

Question

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,

Options:

(A) $56x = 9y$

(B) $29x = 5y$

(C) $45x = 7y$

(D) $21x = 4y$

Sigma Percentile · Accepted Answer

### The Philosophy of Selection

Imagine you are a director casting a play. You have a set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Your goal is to form a 9-digit number.

For $x$, the condition is that exactly one digit repeats twice. This means we have a 'VIP' digit. First, we must choose this VIP from our set of 9, which can be done in $\binom{9}{1}$ ways.

Now, we have used up 2 of our 9 slots. We need 7 more digits to fill the remaining slots, and they must be distinct. We have 8 digits left in our set, so we choose 7 of them: $\binom{8}{7}$.

The number of ways to arrange these 9 items, where 2 are identical, is given by the permutation formula:

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

This simplifies beautifully to:

$$x = 9 	imes 8 	imes \frac{9!}{2} = 36 	imes 9!$$

### The Complexity of $y$

Now, let us elevate the challenge. For $y$, we have two pairs of repeating digits. We are casting two VIPs.

We choose 2 digits from our set of 9 to be our repeating pairs: $\binom{9}{2}$ ways. These two pairs occupy 4 slots. We still need 5 more digits to fill the remaining 5 slots.

We have 7 digits left in our set, so we choose 5: $\binom{7}{5}$. Now, we have 9 digits to arrange: two pairs and five unique ones. The arrangement count is $\frac{9!}{2!2!}$.

Multiplying these, we get:

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

This simplifies to:

$$y = 36 	imes 21 	imes \frac{9!}{4} = 189 	imes 9!$$

### The Grand Finale

We have our values for $x$ and $y$. The beauty of this problem lies in the ratio. When we calculate $\frac{x}{y}$, the $9!$ terms cancel out entirely.

We are left with:

$$\frac{x}{y} = \frac{36 	imes 9!}{189 	imes 9!} = \frac{36}{189}$$

Dividing both the numerator and the denominator by 9, we get:

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

Cross-multiplying gives us the final relationship:

**$21x = 4y$**

By keeping the factorials intact until the very end, we avoided messy arithmetic and let the structure of the problem reveal the answer. Remember, in JEE, the math is not just about calculation; it is about finding the most elegant path to the truth.

Select Answer:

Visualized Solution

Analyzing the Set $S$

Selecting the Repeating Digit for $x$

Selecting Distinct Digits for $x$

Arranging Digits for $x$

Total Count for $x$

Selecting Repeating Digits for $y$

Selecting Distinct Digits for $y$

Arranging Digits for $y$

Total Count for $y$

Setting up the Ratio $\frac{x}{y}$

Simplifying the Fraction

Final Relation

The Sigma Insight: Linear Permutations

The Philosophy of Selection

The Complexity of $y$

The Grand Finale

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