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20 Jan 2024, 09:30
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D
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Difficulty:
25%
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Question Stats:
78% (01:52) correct
22%
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wrong
based on 79
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How many positive n-digit integers, where n is greater than 1, can be formed using nonzero digits such that no two adjacent digits are the same?
A. n^8
B. n^9
C. 8^n
D. 9*8^(n-1)
E. 9^n
A. n^8
B. n^9
C. 8^n
D. 9*8^(n-1)
E. 9^n
22 Jan 2024, 01:30
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How many positive \(n\)-digit integers, where \(n\) is greater than 1, can be formed using nonzero digits such that no two adjacent digits are the same?
A. \(n^8\)
B. \(n^9\)
C. \(8^n\)
D. \(9*8^{(n-1)}\)
E. \(9^n\)
The first digit can take 9 values, from 1 to 9, inclusive.
The next digit can take 8 values, any but the one used for the first digit.
The following digit can also take 8 values, any but the one used for the second digit.
The subsequent digit can also take 8 values, any but the one used for the third digit.
...
And finally, the last, \(n_{th}\) digit can take 8 values, any but the one used for the second to last digit.
Hence, we have 9 options for the first digit and 8 options for each of the remaining \(n-1\) digits, resulting in a total of \(9*8^{(n-1)}\) numbers.
Answer: D
A. \(n^8\)
B. \(n^9\)
C. \(8^n\)
D. \(9*8^{(n-1)}\)
E. \(9^n\)
The first digit can take 9 values, from 1 to 9, inclusive.
The next digit can take 8 values, any but the one used for the first digit.
The following digit can also take 8 values, any but the one used for the second digit.
The subsequent digit can also take 8 values, any but the one used for the third digit.
...
And finally, the last, \(n_{th}\) digit can take 8 values, any but the one used for the second to last digit.
Hence, we have 9 options for the first digit and 8 options for each of the remaining \(n-1\) digits, resulting in a total of \(9*8^{(n-1)}\) numbers.
Answer: D
General Discussion
20 Jan 2024, 09:42
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Bunuel
Let's assume n = 2
_ _
- The first place can be filled in 9 ways (any digit between 1 and 9, both inclusive).
- The second place can be filled in 8 ways (any digit between 1 and 9, both inclusive, except the one used in the first place.
- Number of ways = 9 * 8 = \(9 * 8^1\)
Let's assume n = 3
_ _ _
- The first place can be filled in 9 ways (any digit between 1 and 9, both inclusive).
- The second place can be filled in 8 ways (any digit between 1 and 9, both inclusive, except the one used in the first place.
- The third place can be filled in 8 ways (any digit between 1 and 9, both inclusive, except the one used in the second place.
- Number of ways = 9 * 8 * 8 = \(9 * 8^2\)
Hence, we can infer that the pattern for n-digit integer is ⇒ \(9 * 8^{n-1}\)
Option D
