Four digit numbers are formed using the digits 1,2,3,4 (repetition is allowed). The number of such four digit numbers divisible by 11 is- (1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places. But how do I use this to get the answer?
On
Let $[abcd]$ be the number. Then $[abcd] \equiv [ab]+[cd] \pmod{99}$ which implies $[abcd] \equiv [ab]+[cd] \pmod{11}$. Because $a,b,c,d \in \{1,2,3,4\}$, $[abcd] \equiv [(a+c)(b+d)] \pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
\begin{array}{|r|r|r|} \text{sum} &\text{ac and bd events} & \text{counts} \\ \hline 2 & 11 & 1\\ 3 & 12, \ 21 & 4\\ 4 & 13, \ 22, \ 31 & 9 \\ 5 & 14, \ 23, \ 32, \ 41 & 16\\ 6 & 24, \ 33, \ 42 & 9\\ 7 & 34, \ 43 & 4\\ 8 & 44 & 1 \\ \hline \text{total} && 44 \end{array}
On
Alternatively, consider permutations of numbers made of one, two, three and four digits.
There is one number made of single number $1$, which is $1111$. Total number is: $$1\cdot {4\choose 1}=4.$$
There are four numbers made of two digits $1$ and $2$, which are $1122,1221,2112,2211$. Total number is: $$4\cdot {4\choose 2}=24.$$
There are four numbers made of three digits $1,2,3$, which are $1232,2123,2321,3212$. And there are two possible cases: $1,2,3$ and $2,3,4$. Total number is: $$4\cdot 2=8.$$
There are eight numbers made of four digits $1,2,3,4$, which are $1234,1342,2134,2431,3241,3421,4213,4312$. Total number is: $$8\cdot 1=8.$$
In conclusion, the grand total is: $$4+24+8+8=44.$$
Let $abcd$ be the number.
If $a=b$ then $c=d$. There are $4*4=16$ ways that can occur. (Four options for $a$ and four options for $b$).
If $a=b\pm 1$ then $c=d \mp1$. And there are $2*3*3=18$ ways this can occur. (Two choices whether $a > b$ or $b > a$ and three choices from $1,2,3,4$ that are one apart.
If $a = b\pm 2$ then $c = d\mp 2$ and there $ 2*2*2 = 8$ ways.
And if $a = b\pm 4$ then either $a = 1;b=4;c=4;d=1$ or $a=4;b=1;c=1;d=4$. $2$ ways.
So $16 + 18 + 8 + 2 = 44$ ways.