Question

Consider the set $S=\{0,1,2,3,\cdots ,9\}.$ Let $m$ denotes the number of ways the two numbers $a,b$ with replacement chosen from $S$ such that $|a-b|$ $>6$ and $n$ denotes the number of $10$ digit numbers that can be formed using each and every digit of $S,$ which are divisible by $2$ but not by $3,$ then which of the following is/are correct?

• A. $m+n=12$
• B. $m+n=16$
• C. $m-n=12$
• D. $m-n=8$

Answer 1

Answer: (A), (C)

The correct options are
A $m+n=12$
C $m-n=12$

For $m,|a-b|>6$
$\Rightarrow |a-b|\ge 7$

$\begin{array}{cc}a& b\\ 0& 7,8,9\to 3\text{ways}\\ 1& 8,9\to 2\text{ways}\\ 2& 9\to 1\text{way}\\ 7& 0\to 1\text{way}\\ 8& 0,1\to 2\text{ways}\\ 9& 0,1,2\to 3\text{ways}\end{array}$
Number of ways $m=2(1+2+3)=12$ ways

$S=\{0,1,2,3,\cdots ,9\}$
Sum of the digits is $\frac{9\times 10}{2}=45$ which is divisible by $3.$
So, every $10$ digit number is always divisible by $3.$
$\therefore n=0$