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You are given the $4 \times 4$ grid below. Find the number of ways of placing $8$ counters in the squares (at most one counter per square), so that each row contains exactly two counters, and each column contains exactly two counters.
 Nov 22, 2024
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Let's analyze the problem step by step:

 

Understanding the Problem:

 

We have a 4x4 grid.

 

We need to place 8 counters.

 

Each row and column must have exactly 2 counters.

 

Approach: We can solve this problem using a combinatorial approach. We'll break it down into smaller subproblems.

 

Choosing Rows and Columns:

 

We need to choose 2 rows out of 4 to place counters. This can be done in C(4,2) ways.

 

Similarly, we need to choose 2 columns out of 4. This can also be done in C(4,2) ways.

 

Placing Counters:

 

Once we've chosen the rows and columns, we need to place the counters within those specific rows and columns.

 

For each chosen row, there are C(4,2) ways to choose the two columns where the counters will go.

 

Similarly, for each chosen column, there are C(4,2) ways to choose the two rows.

 

Calculating the Total Number of Ways:

 

Multiplying the number of ways for each step, we get: C(4,2) * C(4,2) * C(4,2) * C(4,2)

 

Calculating the Combinations:

 

C(4,2) = 4! / (2! * 2!) = 6

 

Final Calculation:

 

Total number of ways = 6 * 6 * 6 * 6 = 1296

 

Therefore, there are 1296 ways to place the 8 counters in the 4x4 grid, satisfying the given conditions.

 Nov 22, 2024

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