Let the digits a, b, c be in A.P. Nine-digit | vedclass

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(D) The total number of arrangements of the digits $a, a, a, b, b, b, c, c, c$ is $\frac{9!}{3!3!3!} = 1680$.We want to find the number of arrangement

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$1260$

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(D) The total number of arrangements of the digits $a, a, a, b, b, b, c, c, c$ is $\frac{9!}{3!3!3!} = 1680$.We want to find the number of arrangements where at least one triplet of consecutive digits forms an $A.P.$Since $a, b, c$ are in $A.P.$,the possible triplets in $A.P.$ are $(a, b, c)$ and $(c, b, a)$.Using the principle of inclusion-exclusion or complementary counting,we calculate the arrangements containing at least one such sequence.The number of such nine-digit numbers is $1260$.

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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