Question

The number of ways in which we can choose 2 distinct integers from 1 to 100 such that the difference between them is less than 10 is

• A. ${100}_{C_2}-{90}_{C_2}$
• B. ${100}_{C_{98}}-{90}_{C_{88}}$
• C. ${100}_{C_2}-{90}_{C_{88}}$
  
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A

${100}_{C_2}-{90}_{C_2}$

B

${100}_{C_{98}}-{90}_{C_{88}}$

C

${100}_{C_2}-{90}_{C_{88}}$

Let the chosen integers be $x_1$ and $x_2$

Let there be a integer before $x_1$, b integer between $x_1$ and $x_2$ and c integer after $x_2$.

$\therefore a+b+c=98$. Where $a\ge 0,b\ge 10,c\ge 0$

Now if we consider the choices where difference is at least 10, then the number of solution is ${}^{88+3–1}{C}_{3–1}={90}_{C_2}$

$\therefore$The number of ways in which b is less than 10 is ${100}_{C_2}–{90}_{C_2}$ which is equal to (A), (B) and (C) option.