JEE Mathematics 2026 Question with Solution

Given: Two numbers first $a$ and then $b$ are selected from the first $100$ natural numbers without replacement.

Find: The value of $m+n$ when the probability that $a-b \ge 10$ is $\frac{m}{n}$ in lowest terms.

Since order matters, the total number of outcomes is

$$100 \times 99 = 9900$$

For favorable outcomes, we need

$$a \ge b + 10$$

Now count possible values of $a$ for each value of $b$:

• If $b=1$, then $a$ can be $11,12,\dots,100$, giving $90$ values.
• If $b=2$, then $a$ can be $12,13,\dots,100$, giving $89$ values.
• Continuing similarly,
• If $b=90$, then $a=100$, giving $1$ value.

So the total favorable outcomes are

$$90+89+\cdots+1 = 1+2+\cdots+90 = \frac{90 \times 91}{2} = 4095$$

Therefore, the required probability is

$$P = \frac{4095}{9900}$$

Divide numerator and denominator by $45$:

$$P = \frac{91}{220}$$

Hence,

$$m=91, \quad n=220$$

So,

$$m+n = 91+220 = 311$$

Therefore, the value of $m+n$ is $311$.

Given: Ordered selection without replacement from $1$ to $100$.

Find: Number corresponding to $m+n$.

The hint is that for ordered selection without replacement, sample space size is

$$n(n-1)$$

So here,

$$100 \cdot 99 = 9900$$

We want ordered pairs $(a,b)$ satisfying

$$a-b \ge 10$$

This means for a fixed $b$, the value of $a$ must start from $b+10$ and go up to $100$.

Thus the number of valid $a$ for a given $b$ is

$$100-(b+10)+1 = 91-b$$

This is positive for

$$b=1,2,\dots,90$$

Hence favorable outcomes are

$$\sum_{b=1}^{90} (91-b) = \sum_{k=1}^{90} k = \frac{90 \cdot 91}{2} = 4095$$

So,

$$\text{Probability} = \frac{4095}{9900} = \frac{91}{220}$$

with $\gcd(91,220)=1$.

Therefore,

$$m+n = 91+220 = 311$$

The final answer is $311$.