You are given the 4 x 4 grid below.
(a) 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.
(b) Find the number of ways of placing 12 counters in the squares (at most one counter per square), so that each column contains exactly three counters.
(a) There are 70 ways of placing 8 counters in the squares so that each row contains exactly two counters.
(b) There are 495 ways of placing 12 counters in the squares so that each column contains exactly three counters.
To solve the given problems, you can use a technique called permutation and combination. Let's solve each part one by one.
(a) Find the number of ways of placing 8 counters in the squares so that each row contains exactly two counters.
To solve this, we can break it down into four cases:
Case 1: When all four rows have exactly 2 counters each.
In this case, you need to find the number of ways to arrange 8 counters in a 4 x 2 grid. The number of ways to do this is given by the formula "8 choose 2," which is denoted as C(8,2) or written as 8C2. Evaluating this gives:
C(8,2) = 8! / (2! * (8-2)!) = 8! / (2! * 6!) = 28
Case 2: When three rows have exactly 2 counters each, and one row has 0 counters.
In this case, you need to find the number of ways to arrange 8 counters in a 4 x 2 grid with one empty row. The number of ways to do this is also given by 8C2, as we have to choose 2 positions for the counters. However, since we have 4 possible rows to leave empty, we need to multiply the result by 4. So, the total number of ways is:
4 * C(8,2) = 4 * 28 = 112
Case 3: When two rows have exactly 2 counters each, and two rows have 1 counter each.
In this case, you need to find the number of ways to arrange 8 counters in a 4 x 2 grid with two rows having one counter each. First, choose two positions for the rows with one counter (2C2). Then, choose two positions for the remaining rows with two counters (2C2). Finally, place the counters in the chosen positions. The total number of ways is:
(2C2) * (2C2) * 2! = 1 * 1 * 2 = 2
Case 4: When one row has exactly 2 counters, and three rows have 1 counter each.
In this case, you need to find the number of ways to arrange 8 counters in a 4 x 2 grid with one row having two counters. First, choose one position for the row with two counters (1C1). Then, choose three positions for the remaining rows with one counter each (3C3). Finally, place the counters in the chosen positions. The total number of ways is:
(1C1) * (3C3) * 3! = 1 * 1 * 6 = 6
Now, add up the results from all four cases to get the total number of ways:
Total = Case 1 + Case 2 + Case 3 + Case 4 = 28 + 112 + 2 + 6 = 148
Therefore, there are 148 different ways to place 8 counters in the given 4 x 4 grid such that each row contains 2 counters.
(b) Find the number of ways of placing 12 counters in the squares so that each column contains exactly 3 counters.
To solve this, we can again break it down into multiple cases:
Case 1: When all four columns have exactly 3 counters each.
In this case, you need to find the number of ways to arrange 12 counters in a 3 x 4 grid. The number of ways to do this is given by the formula "12 choose 3," which is denoted as C(12,3) or written as 12C3. Evaluating this gives:
C(12,3) = 12! / (3! * (12-3)!) = 12! / (3! * 9!) = 220
Case 2: When three columns have exactly 3 counters each, and one column has 0 counters.
In this case, you need to find the number of ways to arrange 12 counters in a 3 x 4 grid with one empty column. The number of ways to do this is also given by 12C3, as we have to choose 3 positions for the counters. However, since we have 4 possible columns to leave empty, we need to multiply the result by 4. So, the total number of ways is:
4 * C(12,3) = 4 * 220 = 880
Case 3: When two columns have exactly 3 counters each, and two columns have 1 counter each.
In this case, you need to find the number of ways to arrange 12 counters in a 3 x 4 grid with two columns having one counter each. First, choose two positions for the columns with one counter (2C2). Then, choose two positions for the remaining columns with three counters each (2C2). Finally, place the counters in the chosen positions. The total number of ways is:
(2C2) * (2C2) * 2! = 1 * 1 * 2 = 2
Case 4: When one column has exactly 3 counters, and three columns have 1 counter each.
In this case, you need to find the number of ways to arrange 12 counters in a 3 x 4 grid with one column having three counters. First, choose one position for the column with three counters (1C1). Then, choose three positions for the remaining columns with one counter each (3C3). Finally, place the counters in the chosen positions. The total number of ways is:
(1C1) * (3C3) * 3! = 1 * 1 * 6 = 6
Now, add up the results from all four cases to get the total number of ways:
Total = Case 1 + Case 2 + Case 3 + Case 4 = 220 + 880 + 2 + 6 = 1108
Therefore, there are 1108 different ways to place 12 counters in the given 4 x 4 grid such that each column contains 3 counters.
(a) To find the number of ways of placing 8 counters in the 4x4 grid such that each row contains exactly two counters, we can break down the problem into smaller steps:
Step 1: Select the positions for the first row:
There are 4 choose 2 ways to select 2 positions in the first row. This can be calculated using the binomial coefficient formula: C(4, 2) = 4! / (2! * (4-2)!) = 6.
Step 2: Select the positions for the second row:
Since each row must contain exactly two counters, there are only 2 positions remaining in each row. Therefore, there are no choices to be made for the second row.
Step 3: Select the positions for the third row:
Again, we have 2 positions remaining in each row, so there are no choices to be made.
Step 4: Select the positions for the fourth row:
As in step 1, there are 4 choose 2 ways to select 2 positions in the fourth row: C(4, 2) = 6.
Step 5: Multiply the number of choices for each step together:
Since each step is independent of the others, we can multiply the number of choices for each step together to find the total number of ways.
Total number of ways = 6 * 1 * 1 * 6 = 36.
Therefore, there are 36 ways of placing 8 counters in the 4x4 grid such that each row contains exactly two counters.
(b) To find the number of ways of placing 12 counters in the 4x4 grid such that each column contains exactly three counters, we can again break down the problem into smaller steps:
Step 1: Select the positions for the first column:
There are 4 choose 3 ways to select 3 positions in the first column: C(4, 3) = 4.
Step 2: Select the positions for the second column:
Since each column must contain exactly three counters, there are only 1 position remaining in each column. Therefore, there are no choices to be made for the second column.
Step 3: Select the positions for the third column:
Again, we have 1 position remaining in each column, so there are no choices to be made.
Step 4: Select the positions for the fourth column:
As in step 1, there are 4 choose 3 ways to select 3 positions in the fourth column: C(4, 3) = 4.
Step 5: Multiply the number of choices for each step together:
Since each step is independent of the others, we can multiply the number of choices for each step together to find the total number of ways.
Total number of ways = 4 * 1 * 1 * 4 = 16.
Therefore, there are 16 ways of placing 12 counters in the 4x4 grid such that each column contains exactly three counters.