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The number of...

Question

The number of 5 digited numbers in which no two consecutive digits are identical is
A) 9(81^2) B) (64)^3 C) 9(64)^2 D) 120

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Solution


We can choose the first digit any of 9 ways, 1,2,3,4,5,6,7,8 or 9
We can then choose the 2^nd digit any of 9 ways, 0 or any of 8 ways not chosen for the 1^st digit
We can then choose the 3^rd digit any of the 9 ways we didn't choose for 2^nd digit.
We can then choose the 4^th digit any of the 9 ways we didn't choose for the 3^rd digit.
We can then choose the 5^th digit any of the 9 ways we didn't choose for the 4^th digit.
So total number =  $9 \times 9 \times 9 \times 9 \times 9 = 9^{5} = 9 {(81)}^{2}$

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