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,
The correct answer is
$21x = 4y$
Understanding the Problem
The problem requires calculating two quantities, $x$ and $y$, related to forming 9-digit numbers using digits from the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. We need to determine the relationship between $x$ and $y$.
- $x$: Counts 9-digit numbers where exactly one digit from $S$ is repeated exactly twice.
- $y$: Counts 9-digit numbers where exactly two digits from $S$ are each repeated exactly twice.
Calculating x: One Digit Repeated Twice
To form a 9-digit number with exactly one digit repeated twice:
- Step 1: Choose the repeated digit. There are $\binom{9}{1}$ ways to choose which digit from $S$ will be repeated.
- Step 2: Choose the other distinct digits. Since one digit is used twice, we need 7 more distinct digits to complete the 9 digits. These must be chosen from the remaining $9 - 1 = 8$ digits in $S$. There are $\binom{8}{7}$ ways.
- Step 3: Arrange the digits. We have a total of 9 digits (the repeated digit twice, and 7 other distinct digits). The number of ways to arrange these 9 digits is given by the formula for permutations with repetitions: $\frac{9!}{2!}$.
The total count for $x$ is:
$x = \binom{9}{1} \times \binom{8}{7} \times \frac{9!}{2!} = 9 \times 8 \times \frac{9!}{2} = 72 \times \frac{9!}{2} = 36 \times 9!$Calculating y: Two Digits Repeated Twice
To form a 9-digit number with exactly two digits repeated twice:
- Step 1: Choose the two digits to be repeated. There are $\binom{9}{2}$ ways to choose which two digits from $S$ will each appear twice.
- Step 2: Choose the other distinct digits. We have used 4 digits (two pairs). We need $9 - 4 = 5$ more distinct digits. These must be chosen from the remaining $9 - 2 = 7$ digits in $S$. There are $\binom{7}{5}$ ways.
- Step 3: Arrange the digits. We have a total of 9 digits (two digits repeated twice, and 5 other distinct digits). The number of ways to arrange these 9 digits is $\frac{9!}{2!2!}$.
The total count for $y$ is:
$y = \binom{9}{2} \times \binom{7}{5} \times \frac{9!}{2!2!} = \left(\frac{9 \times 8}{2}\right) \times \left(\frac{7 \times 6}{2}\right) \times \frac{9!}{4}$ $y = 36 \times 21 \times \frac{9!}{4} = \frac{36 \times 21}{4} \times 9! = 9 \times 21 \times 9! = 189 \times 9!$Verifying the Relationship
We are given options relating $x$ and $y$. Let's test the relationship presented in Option A: $21x = 4y$.
- Substitute the calculated values: $21 \times (36 \times 9!) \stackrel{?}{=} 4 \times (189 \times 9!)$
- Calculate the products: $(21 \times 36) \times 9! = (4 \times 189) \times 9!$ $756 \times 9! = 756 \times 9!$
The equality holds true, confirming that $21x = 4y$.
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