Question

Total number of $4$ digit numbers that can be formed using the digits $0,1,2,...9$, when

• A. the repetition is allowed is ${10}^4$
• B. the repetition is not allowed is $9\times {}^9{P}_3$
• C. the number is even and the repetition is allowed is $4500$
• D. the number is divisible by $5$ and repetition is not allowed is $952$

Answer 1

Answer: (B), (C), (D)

the number is divisible by $5$ and repetition is not allowed is $952$

Since, $0$ cannot appear at thousand's place it can be filled by numbers from $1$ to $9$ only i.e, $9$ ways

When repetition is allowed.
The rest $3$ places can be filled in ${10}^3$ ways.
Hence, the total number of $4$ digit numbers that can be formed is : $9\times {10}^3$

When repetition is not allowed.
The rest $3$ places can be filled in${}^9{P}_3$ ways.
Hence, the total number of $4$ digit numbers that can be formed is : $9\times {}^9{P}_3$

When the number is even and repetition is allowed.
The last digit or units digit can be filled by $\{0,2,4,6,8\}$i.e. $5$ways, rest $2$ places can be filled in ${10}^2$ ways.
Hence, the total number of $4$ digit numbers that can be formed is : $9\times {10}^2\times 5=4500$

When the number is divisible by $5$ and the repetition is not allowed is
$\left(1\right)$ If the last digit is $0$, the rest $3$ places can be filled in${}^9{P}_3$ ways.
$\left(2\right)$ If the last digit is $5$, the first digit can be filled in $8$ ways and the rest $2$ places can be filled in${}^8{P}_2$ ways.
Hence, the total number of $4$ digit numbers that can be formed is $={}^9{P}_3+8\times {}^8{P}_2=952$