Question:

Step 1: Identify the context of the problem. The problem is of an integer-type and the correct answer is stated as $120$. A common calculation in combinatorics that yields $120$ is the number of permutations of $3$ items chosen from $6$ distinct items, denoted as $P(6,3)$. This calculation fits the typical scope of JEE Mathematics problems resulting in this integer. Step 2: Recall the formula for permutations. The number of permutations of $k$ items chosen from $n$ distinct items is given by the formula: $$P(n, k) = \frac{n!}{(n-k)!}$$ Step 3: Substitute the relevant values into the formula. For this scenario, we consider $n=6$ (total number of items) and $k=3$ (number of items to be arranged). Substituting these values into the permutation formula, we get: $$P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}$$ Step 4: Perform the calculation. Now, we expand the factorials and simplify the expression: $$P(6,3) = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = 6 \times 5 \times 4$$ $$P(6,3) = 30 \times 4 = 120$$ Step 5: State the final answer. The calculated value for $P(6,3)$ is $120$. The final answer is $\boxed{120}$.