Each of the digits $\; 1, \; 1, \; 2, \; 3, \; 3, \; 4 \;$ is written on a separate card. The six cards are then laid out in a row to form a 6-digit number.
- How many distinct 6-digit numbers are there?
- How many of these 6-digit numbers are even?
- How many of these 6-digit numbers are divisible by 4?
- Of the given digits, there are two 1's and two 3's.
$\therefore \;$ Number of distinct 6-digit numbers $= \dfrac{6!}{2! \times 2!} = \dfrac{6 \times 5 \times 4 \times 3}{2} = 180$ - The unit's place can either be a $2$ or a $4$.
$\therefore \;$ The unit's place can be selected in $2$ ways.
Then the remaining five places can be selected in $\dfrac{5!}{2! \times 2!} = \dfrac{5 \times 4 \times 3}{2} = 30$ ways
$\therefore \;$ Number of 6-digit numbers which are even $= 2 \times 30 = 60$ numbers - Required 6-digit numbers which are divisible by 4 are $\; X \; X \; X \; X \; 1 \; 2$ $\;\;$ OR $\;\;$ $\; X \; X \; X \; X \; 2 \; 4$
When numbers are of the form $\; X \; X \; X \; X \; 1 \; 2$:
The unit's place can be selected in $1$ way;
the ten's place can be selected from the two 1's in $2$ ways;
the remaining four places can be selected from the remaining digits $\; 1, \; 3, \; 3, \; 4 \;$ in
$\dfrac{4!}{2!} = \dfrac{4 \times 3 \times 2!}{2!} = 12$ ways
$\therefore \;$ Numbers of the form $\; X \; X \; X \; X \; 1 \; 2$ can be selected in $2 \times 12 = 24$ ways
When numbers are of the form $\; X \; X \; X \; X \; 2 \; 4$:
The unit's and the ten's place can be selected in $1$ way each.
The remaining four places can be selected from the remaining digits $\; 1, \; 1, \; 3, \; 3 \;$ in
$\dfrac{4!}{2! \times 2!} = \dfrac{4 \times 3 \times 2!}{2! \times 2} = 6$ ways
$\therefore \;$ Numbers of the form $\; X \; X \; X \; X \; 2 \; 4$ can be selected in $1 \times 1 \times 6 = 6$ ways
$\therefore \;$ Number of 6-digit numbers which are divisible by 4 $= 24 + 6 = 30$ numbers