Stuart has some 2-ounce, 3-ounce, and 4-ounce weights. How many different ways can Stuart combine the weights to make a total of 12 ounces?
start with as many big weights as you can.
4+4+4
4+4+2+2
since 4+2 = 3+3, you can now go to
4+3+3+2
now start working to eliminate the last 4. There aren't very many more ways to do it. Eventually you will get all the way down to
2+2+2+2+2+2
To find the different ways Stuart can combine the weights to make a total of 12 ounces, we can use a recursive approach or dynamic programming.
One way to solve this problem is by using dynamic programming with an array to keep track of the ways to reach each weight up to 12 ounces.
1. Create an array of 13 elements (0 to 12) and initialize all elements to 0. This array will represent the number of different ways to make each weight.
2. Initialize the first element of the array to 1, since there is only one way to make 0 ounces (by not using any weight).
3. Iterate over each weight in the given weights collection (2, 3, and 4 ounces).
4. For each weight, iterate over the target weights (from the weight being considered up to 12 ounces) and update the array element at that weight if it is not greater than 12.
5. Update the array element by adding the number of ways to reach the remaining weight. The remaining weight is the target weight minus the current weight.
6. Finally, the value at the array element with index 12 will represent the number of different ways to make a total of 12 ounces using the given weights.
Let's go ahead and calculate the number of different ways Stuart can combine the weights to make a total of 12 ounces using this approach.
Initialization:
ways = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
Iteration 1:
weight = 2
For weights from 2 to 12: ways[2] = ways[2] + ways[0] = 1, ways[4] = ways[4] + ways[2] = 0, ways[6] = ways[6] + ways[4] = 0, ways[8] = ways[8] + ways[6] = 0, ways[10] = ways[10] + ways[8] = 0, ways[12] = ways[12] + ways[10] = 0
Current state: ways = [1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
Iteration 2:
weight = 3
For weights from 3 to 12: ways[3] = ways[3] + ways[1] = 0, ways[5] = ways[5] + ways[3] = 1, ways[7] = ways[7] + ways[5] = 1, ways[9] = ways[9] + ways[7] = 1, ways[11] = ways[11] + ways[9] = 1
Current state: ways = [1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0]
Iteration 3:
weight = 4
For weights from 4 to 12: ways[4] = ways[4] + ways[0] = 1, ways[6] = ways[6] + ways[4] = 1, ways[8] = ways[8] + ways[6] = 1, ways[10] = ways[10] + ways[8] = 1, ways[12] = ways[12] + ways[10] = 1
Current state: ways = [1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
The final value at ways[12] is 1, which means there is one way to combine the weights to make a total of 12 ounces.
Therefore, Stuart can combine the given weights in one way to make a total of 12 ounces.