In how many ways can cells in a 3x3 grid be shaded such that each row and column has exactly one shaded cell?

There are 6 ways. This is equivalent to counting the number of permutations of 3 elements, which is 3! = 3 × 2 × 1 = 6.

Why is the answer to the 3x3 grid shading problem 6?

Each valid shading corresponds to placing one shaded cell in each row such that no two cells share the same column. This is exactly the number of ways to arrange 3 distinct columns across 3 rows, which is 3! = 6.

What concept is tested in GATE Mining Engineering 2023 Question 3?

This question tests the concept of permutations in combinatorics. The 3x3 grid shading problem is a classic application of counting arrangements where each row and column must have exactly one selection.

What is 3 factorial (3!)?

3 factorial, written as 3!, equals 3 × 2 × 1 = 6. It represents the total number of ways to arrange 3 distinct items in a sequence.

3x3 Grid Shading Ways: One Cell Per Row & Column | GATE Mining Engineering 2023 Q3

Mining Engineering > GATE 2023 > Aptitude

In how many ways can cells in a 3x3 grid be shaded, such that each row and each column have exactly one shaded cell? An example of one valid shading is shown.

Correct : d

The correct answer is 6, which is option D.
In a 3×3 grid, you need to place exactly one shaded cell in each row and each column. For the first row, you have 3 columns to choose from. Once you pick one, the second row has only 2 remaining columns available, and the third row is left with just 1 choice.
So the total number of ways = 3 × 2 × 1 = 3! = 6.
This is a straightforward permutation problem. Each valid shading arrangement is nothing but a permutation of the column positions {1, 2, 3} across the three rows and there are exactly 6 such permutations.

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