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Q.

Let the digits a, b, c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?

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answer is 1260.

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Detailed Solution

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$Digits are in A.P. at least once p o s s i b l e w a y s a b c o r c a b$
required number of numbers = $\frac{{}^{7}C_{1} \times 2 \times 6!}{2! 2! 2!} = \frac{7!}{4!}$
= 1260

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