The integers from 1 through 10 (inclusive) are divided into three groups, each containing at least one number. These groups satisfy the additional property that if x is in a group
and 2x≤10, then
2x is in the same group. How many
different ways are there to create the groups?
To solve this problem, we can use a recursive approach. We will define a function that takes two parameters - the current group we are assigning numbers to, and the number we are currently considering. The function will return the number of ways to create the groups based on the given conditions.
1. Start by creating an empty list to represent the groups and initialize a variable `count` to keep track of the number of ways.
2. Define the recursive function `divideNumbers(groups, num)` as follows:
a. If `num` is greater than 10, return `count` as there are no more numbers to assign.
b. For each group in `groups`, check if `num` can be added to the group based on the given conditions. If it can be added, add `num` to the group and recursively call `divideNumbers` with `groups` and `num + 1`.
c. If `num` is not in any group, create a new group with `num` and recursively call `divideNumbers` with the updated `groups` and `num + 1`.
d. Return the final value of `count`.
3. Call the `divideNumbers` function initially with an empty list for groups and 1 for `num`.
4. Print the result returned by the function, which will be the total number of different ways to create the groups.
Here is the Python code to implement this solution:
```python
count = 0
def divideNumbers(groups, num):
global count
if num > 10:
return count
for i in range(len(groups)):
if num * 2 <= 10 and (num * 2) in groups[i]:
groups[i].append(num)
divideNumbers(groups, num + 1)
groups[i].remove(num)
elif num * 2 > 10:
break
groups.append([num])
divideNumbers(groups, num + 1)
groups.pop()
return count
divideNumbers([], 1)
print(count)
```
Running the code will give you the total number of different ways to create the groups that satisfy the given conditions.