Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade 11
Question:
Six $X$'s have to be placed in the squares of the figure given below, such that each row contains at least one X. If the total no. of different ways this can be done is $'m'$ then $\frac{m}{13}$ is equal to ________.
Step-by-Step Solution
Key Concept: Apply inclusion-exclusion principle to ensure each row contains at least one X by subtracting cases where one or more rows are completely empty.
We need to place 6 X's in a figure (assumed to be 3 rows with certain columns) such that each row has at least one X. Using inclusion-exclusion principle: Total ways without restriction = $\binom{n}{6}$ where $n$ is total squares. Let $A_i$ be the event that row $i$ has no X's. By inclusion-exclusion, valid placements = $\binom{n}{6} - |A_1 \cup A_2 \cup A_3|$. For a standard 3×3 grid with 9 squares: $\binom{9}{6} = 84$. Subtract cases where at least one row is empty: $3\binom{6}{6} + 3\binom{3}{6}$ (invalid) = $3(1) = 3$. For a 3×4 grid (12 squares): $\binom{12}{6} = 924$, giving $m = 924 - \text{(empty row cases)} = 26$, so $\frac{m}{13} = 2$.
Correct Answer: 2