Question ID: #919
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,
- (1) $29x=5y$
- (2) $45x=7y$
- (3) $21x=4y$
- (4) $56x=9y$
Solution:
**Calculate x:**
We need to form a 9-digit number where one digit is repeated twice, and the rest are distinct.
Total digits required = 8 distinct digits from set S (since 1 is used twice, plus 7 others = 9 positions).
1. Select the digit to be repeated from 9 digits: ${}^9C_1$.
2. Select the remaining 7 digits from the remaining 8 digits: ${}^8C_7$.
3. Arrange these 9 digits (where 2 are identical): $\frac{9!}{2!}$.
$$x = {}^9C_1 \times {}^8C_7 \times \frac{9!}{2!} = 9 \times 8 \times \frac{9!}{2} = 36 \times 9!$$
**Calculate y:**
We need to form a 9-digit number where two digits are repeated twice each, and the rest are distinct.
Total digits required = 7 distinct digits from set S (2 repeated + 5 others = 7 distinct digits filling 9 positions).
1. Select the 2 digits to be repeated from 9 digits: ${}^9C_2$.
2. Select the remaining 5 digits from the remaining 7 digits: ${}^7C_5$.
3. Arrange these 9 digits (where two pairs are identical): $\frac{9!}{2!2!}$.
$$y = {}^9C_2 \times {}^7C_5 \times \frac{9!}{2!2!} = 36 \times 21 \times \frac{9!}{4} = 9 \times 21 \times 9! = 189 \times 9!$$
**Find relation:**
$$\frac{x}{y} = \frac{36 \times 9!}{189 \times 9!} = \frac{36}{189}$$
Divide numerator and denominator by 9:
$$\frac{x}{y} = \frac{4}{21}$$
$$21x = 4y$$
Ans. (3)
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