Question

How many different $4$-digit numbers are possible to construct using the digits $1,3,4,6,7,8$ that satisfy all these conditions.
$1.$ the number is between $3300$ and $7200$
$2.$ the number is even
$3.$ the number has no repeated digit.

• A. 84
• B. 84.0
• C. 84.00

Answer 1

Answer: (A), (B), (C)

Odd digits $-1,3,7.$ Even digits $-4,6,8$
$1^{st}$ place can be filled by $3,4,6,7$

Case $\text{I}:$ $3$ is at $1^{st}$ place.
$\begin{array}{cccc}3& & & \end{array}$
Since number is even, $4^{th}$ place can be filled in $3$ ways and $2^{nd}$ place in $3$ ways.
Number of ways $=1\times 3\times 3\times 3=27$

Case $\text{II}:$ $4$ is at $1^{st}$ place.
$\begin{array}{cccc}4& & & \end{array}$
Number of ways $=1\times 4\times 3\times 2=24$

Case $\text{III}:$ $6$ is at $1^{st}$ place.
$\begin{array}{cccc}6& & & \end{array}$
Number of ways $=1\times 4\times 3\times 2=24$

Case $\text{IV}:$ $7$ is at $1^{st}$ place.
$\begin{array}{cccc}7& & & \end{array}$
Number of ways $=1\times 1\times 3\times 3=9$

Hence, total number of ways
$=27+24+24+9=84$ ways