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?

Question

DivineJEE · Accepted Answer

There are only two ways to write digits $a, b, c$ , be in A.P.\n\n

\n\n\begin{aligned}\n& \mathrm{I}^{\mathrm{It}} a b c={ }^2 C_1 \\\n& \mathrm{In}^{\text {nd }} \\\n& -\n\n\end{aligned}\n\n

\n \nHere, 9 places to put

a b c

or

b c a

but there will be only 7 possible ways A.P. to choose three consecutive numbers are i.e.,\n \n

\n\n123,234,345,456,567,678,789={ }^7 C_1\n\n

\n \nNow, we have left only 6 places where

a, b, c

three such that three consecutive digits are in A.P.\n\n

\n\n\begin{aligned}\n& \Rightarrow \frac{6!}{2!2!2!} \\\n& \text { Required number }={ }^2 C_1 \cdot{ }^7 C_1 \cdot \frac{6!}{2!2!2!} \\\n& =\frac{2 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{2 \times 2 \times 2}=1260\n\n\end{aligned}\n\n

\n

Explanation

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