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22 Jan 2024, 01:29
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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, 01:29
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Official Solution:
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
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
