Question

Number of $4$ digit numbers using digits $0,1,2,3,4,5$ which are divisble by $11$, when each digit is used at most once is

• A. $40$
• B. $24$
• C. $48$
• D. $96$

Answer 1

The correct option is C $48$

Number is divisible by $11$ if $a_1-a_2+a_3-a_4+\cdots \cdots +(-1{)}^n{a}_n$is multiple of $11$

Let the required $4$ digit number be $a_1{a}_2{a}_3{a}_4,a_1\ne 0$
for four digit numbers $a_1-a_2+a_3-a_4=11k$
$\Rightarrow a_1+a_3=a_2+a_4+11k$

Since here sum of any two digits is less than $11$

so, only possiblity $a_1+a_3=a_2+a_4$

$\therefore$ to satsfy the required condition the pair of digits in even and odd positions are $=\left(\right(3,0),(1,2\left)\right)$ or $\left(\right(4,0),(1,3\left)\right)$ or $\left(\right(5,0),(2,3\left)\right)$ or $\left(\right(5,0),(1,4\left)\right)$ or $\left(\right(2,3),(1,4\left)\right)$ or $\left(\right(1,5),(2,4\left)\right)$ or $\left(\right(5,2),(4,3\left)\right)$

Now if $(a_1,a_3)=(3,0)$ and $(a_2,a_4)=(1,2)$ then while forming the numbers $(a_2,a_4)$ can interchange in $2$ possible ways
and if $(a_1,a_3)=(1,2)$ and $(a_2,a_4)=(3,0)$ then while forming the numbers $(a_1,a_3),(a_2,a_4)$ can interchange. so $4$ ways are possible
Hence when $0$ is included $6$ numbers can be formed

similarly for $(a_1,a_2)\equiv (2,3),(1,5),(5,2)$ total $8$ ways are possible for each of the three
hence, total numbers$=4\times 6+3\times 8=48$