Mathematics - Permutation Combination Question with Solution | TestHub

Three numbers $a, b, c$ are in A.P. ad they are used to making $9-$ digit number using each digit thrice such that atleast three consecutive digits are in A.P., So, $c,b,a$ will also be in A.P.

Now, we have $7$ position to put $\left(a,b,c\right) \mathrm{or} \left(c,b,a\right)$ as $\underset{7 \mathrm{positions}}{\underbrace{\_ \_ \_ \_ \_ \_ \_}} \_ \_$.

Number of such numbers is

$={}^{2}C_1·{}^{7}C_1·\frac{6!}{2!·2!·2!}$

$=1260$