Question

The number of ways to select two different natural numbers which are less than or equal to 100 and differ by at most 10 is

• A. 1034
• B. 945
• C. 935
• D. 1122

Answer 1

The correct option is B

945

Let the numbers be $N_1$ and $N_2$ such that $N_1<N_2$

Then $N_1$ and $N_2$ divide the set of first 100 natural numbers in three parts :

$x$ numbers are less than $N_1,y$ numbers are between $N_1$ and $N_2$ and $z$ numbers are greater than $N_2$

Since, $x+N_1+y+N_2+z=100$ So, $x+y+z=100-1-1=98$.

Now, given condition is that $N_1$ and $N_2$ differ by at most 10

However, it is better to do the complementary counting.

$\Rightarrow y\ge 10$ as a minimum value of 10 ensures difference of $N_1$ and $N_2$ to be more than 10, which serves as the complementary condition.

Total number of ways of selecting two different natural numbers which are less than or equal to 100 is ${}^{100}{C}_2$
Let $y=t+10.$ Then we need number of non-negative integral solution for $x+z+t=88$ viz. ${}^{90}{C}_2$
$\Rightarrow$ Required ways ${=}^{100}{C}_2{-}^{90}{C}_2=945$