Question

Total $4$ digit numbers that can be formed, such that the number should contain exactly one $7$ and exactly two consecutive digits that are identical (other than $7$ and $0$) are

Answer 1

For a $4$ digit number $\to$

One digit will be $7$
Other two digits which are identical and consecutive will be from $1-9$ other than $7$ so there is $8$ possibilities for one (and the other number will be identical to it)
The last digit can be selected from the remaining $8$ digits.
Then they can be arranged in $3!$ ways ($\because$ two digits are consecutive and identical)

Total digits that can be formed will be $1\times (8\times 1)\times 8\times 3!\times \frac{2!}{2!}=384$
Subtracting the cases where there is $0$ in the beginning
Total cases with $0$ in the beginning $=1\times (8\times 1)\times 1\times 2!\times \frac{2!}{2!}=16$
Required numbers will be $=384-16=368$