Let x be the number of 9 digit numbers formed by taking digits from first 9 natural numbers, where only one digit is repeated twice & y be the number of 9 digit numbers formed from first 9 natural numbers, such that exactly 2 digits repeated twice then
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Always perform selection (combinations) first, then arrangement (permutations) to avoid errors in counting problems.
To solve the problem, we need to calculate the values of \(x\) and \(y\), and then check the relationship between them as given in the options.
Calculate \(x\):
\(x\) is the number of 9-digit numbers formed by using the first 9 natural numbers (1 to 9) such that exactly one digit is repeated twice. The rest of the digits are used once.
Select the digit to be repeated: There are 9 choices (digits 1 through 9).
Select 7 other digits from the remaining 8: Choose 7 out of 8 digits. This can be done in \(\binom{8}{7} = 8\) ways.
All digits (8 digits, where one is repeated) can be arranged forming a 9-digit number. The total number of permutations for these 9 digits, where one is repeated twice, is \(\frac{9!}{2!}\).
So, \(x = 9 \times 8 \times \frac{9!}{2!}\).
Calculate \(y\):
\(y\) is the number of 9-digit numbers formed by using the first 9 natural numbers such that exactly two digits each are repeated twice.
Select the two digits to be repeated: There are \(\binom{9}{2} = 36\) ways.
The remaining 5 digits can be chosen from the other 7 digits: \(\binom{7}{5} = 21\) ways.
The 9 digits include 4 repeats (two digits are repeated twice). The total number of permutations for these 9 digits is \(\frac{9!}{2! \times 2!}\).
Hence, \(y = 36 \times 21 \times \frac{9!}{4}\).
Determine the relationship between \(x\) and \(y\):
Given options: We compare \(x\) and \(y\) to the options provided. From previous calculations, we can verify:
\[\frac{x}{y} = \frac{9 \times 8 \times \frac{9!}{2}}{36 \times 21 \times \frac{9!}{4}}\]
Simplifying \(\frac{x}{y} = \frac{21}{4}\), which implies \(21x = 4y\).
