Permutations and Combinations Part 2

Concept:We need to count 9-digit numbers from digits 1-9 with specific repetition patterns.For $x$, one digit repeats twice and all others appear once.For $y$, two digits each repeat twice and the rest appear once.Explanation:For $x$: Choose the repeated digit: ${ }^{9}C_1$ ways.From the remaining 8 digits, choose 7 distinct digits: ${ }^{8}C_7$ ways.Arrange the 9 digits (with one digit repeated twice): $\frac{9!}{2!}$ ways.Thus, $x = { }^{9}C_1 \times { }^{8}C_7 \times \frac{9!}{2!}$.For $y$: Choose the two repeated digits: ${ }^{9}C_2$ ways.From the remaining 7 digits, choose 5 distinct digits: ${ }^{7}C_5$ ways.Arrange the 9 digits (with two digits each repeated twice): $\frac{9!}{2!2!}$ ways.Thus, $y = { }^{9}C_2 \times { }^{7}C_5 \times \frac{9!}{2!2!}$.Divide $x$ by $y$ to find the relation:$\frac{x}{y} = \frac{{}^{9}C_1 \times {}^{8}C_7 \times \frac{1}{2}}{{}^{9}C_2 \times {}^{7}C_5 \times \frac{1}{4}} = \frac{9 \times 8 \times \frac{1}{2}}{36 \times 21 \times \frac{1}{4}} = \frac{36}{189} = \frac{4}{21}$.Therefore, $21x = 4y$.Answer:Option B: $21 x = 4 y$.