You are given the 4x4 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.
I attempted this problem with combinations, but still could not do it. Could anyone that is expert enough to solve this help me?!
The number of ways to place 8 counters in the grid so that each row contains exactly two counters is the same as the number of ways to place 8 counters in the grid so that each column contains exactly two counters. This is because we can simply rotate the grid 90 degrees to get from one case to the other.
So, let's consider the problem of placing 8 counters in the grid so that each column contains exactly two counters. We can think of this as a placement problem, where we have 8 counters and 4 columns, and we want to place the counters in the columns so that no column has more than 2 counters.
There are a few ways to solve this problem. One way is to use casework. We can consider the following cases:
Case 1: 2 counters in one column and 1 counter in each of the other 3 columns.
Case 2: 1 counter in one column and 2 counters in each of the other 3 columns.
Case 3: 1 counter in each of 2 columns and 2 counters in each of the other 2 columns.
For each case, we can count the number of ways to place the counters. For example, in Case 1, there are 4 ways to choose which column will have 2 counters, and then there are 3 ways to choose which of the remaining 3 columns will have the first counter, and then 2 ways to choose which of the remaining 2 columns will have the second counter. So, there are 4×3×2=24 ways to place the counters in Case 1.
Similarly, there are 4×3=12 ways to place the counters in Case 2, and 4×2=8 ways to place the counters in Case 3.
So, the total number of ways to place the counters is 24+12+8=44.
