Question

The number of ways in which we can choose 2 distinct integers from 1 to 100 such that difference between them is at most 10 is
A. ${}^{40}{C_2}$
B. ${}^{70}{C_2}$
C. ${}^{100}{C_2} - {}^{90}{C_2}$
D. None of these

Answer 1

First, we will choose any number between 1 and 100 say $x$. We can choose any number between $x + 1$ and $x + 10$, which gives us 10 choices for each. But if we choose the number to be 91, then we will have nine numbers to choose, that is, 92 to 100 and so on. After calculating the required value, we will consider each option and find the values, using a formula to calculate combinations is ${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\\! \right| }}{{\left. {\underline {\, r \,}}\\! \right| \cdot \left. {\underline {\, {n - r} \,}}\\! \right| }}$, where $n$ is the number of items, and $r$ represents the number of items being chosen.

Complete step by step answer:

We are given that the number of ways in which we can choose 2 distinct integers from 1 to 100 such that the difference between them is at most 10.

Let us choose any number between 1 and 100 say $x$.

You can choose any number between $x + 1$ and $x + 10$, which gives us 10 choices for each.

If we choose the number to be 91, then we will have nine numbers to choose from, that is, 92 to 100.

Similarly, if we choose the number 93, it gives us 8 choices.

So, number 94 gives us 7 and so on, until 99 gives us only 1 choice.

Thus, we will now find the number of ways in which we can choose 2 distinct integers.

$$\Rightarrow \left( {90 \times 10} \right) + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 \\\ \Rightarrow 900 + 45 \\\ \Rightarrow 945 \\\$$

We know that the formula to calculate combinations is ${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\\! \right| }}{{\left. {\underline {\, r \,}}\\! \right| \cdot \left. {\underline {\, {n - r} \,}}\\! \right| }}$, where $n$ is the number of items, and $r$ represents the number of items being chosen.

Consider option A, substituting the values of $n$ and $r$ in the above formula of combinations, we get

$$\Rightarrow \dfrac{{40!}}{{2!\left( {40 - 2} \right)!}} \\\ \Rightarrow \dfrac{{40 \times 39 \times 38!}}{{2!38!}} \\\ \Rightarrow \dfrac{{40 \times 39}}{2} \\\ \Rightarrow 780 \\\$$

Since the number 780 is not equal to 945, option A is incorrect.

Consider option B, substituting the values of $n$ and $r$ in the above formula of combinations, we get

$$\Rightarrow \dfrac{{70!}}{{2!\left( {70 - 2} \right)!}} \\\ \Rightarrow \dfrac{{70 \times 69 \times 68!}}{{2!68!}} \\\ \Rightarrow \dfrac{{70 \times 69}}{2} \\\ \Rightarrow 2415 \\\$$

Since number 2415 is not equal to 945, option B is also incorrect.

Consider option C, substituting the values of $n$ and $r$ in the above formula of combinations, we get

$$\Rightarrow \dfrac{{100!}}{{2!\left( {100 - 2} \right)!}} - \dfrac{{90!}}{{2!\left( {90 - 2} \right)!}} \\\ \Rightarrow \dfrac{{100 \times 99 \times 98!}}{{2!98!}} - \dfrac{{90 \times 89 \times 88!}}{{2!88!}} \\\ \Rightarrow \dfrac{{100 \times 99}}{2} - \dfrac{{90 \times 89}}{2} \\\ \Rightarrow 4950 - 4005 \\\ \Rightarrow 945 \\\$$

Since 945 is equal to 945, option C is correct.

Hence, option C is correct.

Note: The students can make an error while calculating the combinations for each option. Here, students must take care while simplifying the conditions given in the question into the combinations. Since the choices after 90 will keep on decreasing, we must add the choices separately.