## I have 10 variables and I have to fill them with either 1 or -1 , how do I list out all the possible cases?

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I'm working on a math problem dealing with arrangements of Males and Females. There are 5 couples and we know that each couple sit adjacent to each other , every male is seated opposite to a female . If I am to represent male or female with either +1 or -1 and want to find the list of all possible outcomes and print them, how do I go about it?

Do I have to make use of 10 FOR loops? Is there a simpler way in which I could perhaps store the results in an array?

P.S. The math question isnt limited to what I have written here, I want to solve it by myself(sorry) but I would like to understand the issue I have highlighted above. If possible show me a small snippet to point me in the right way.

The most "efficient" or "simple" way to generate all ways to get 10 numbers containing `1` or `-1` depends on the language. Python has a very simple and quick way. Here is an expression that creates a generator that does this. First execute the command

```import itertools
```

to get what you need into your namespace, then the expression is

```itertools.product((1, -1), repeat=10)
```

For example, to work on all those tuples of 10 numbers you could do:

```import intertools

for mytuple in itertools.product((1, -1), repeat=10):
# Process mytuple
```

If you are not familiar with Python, each "tuple" that is created for a cycle of the loop is basically equivalent to an array.

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Check out the Power Set to generate all possible subsets of a set.

There are many ways to generate this, but you stated that you want to figure it out.

This link from math.stackexchange.com seems relevant

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If order doesn't matter, there are eleven solutions: one for each possible number of -1 in the output list. To list these, loop from 0 to 10 (inclusive) and then loop from 1 to 10 (inclusive), first printing a number of -1 equal to the value of the outer counter, then printing 1 until the end of the inner loop.

If order does matter, a simple recursive solution is to start with an empty buffer and position zero. Then, at each level, add -1 and 1, in turn, to the current position, recursively filling the remainer of the buffer by incrementing the position in each call. When the position is equal to the buffer length (hence the buffer is full) you can print the buffer's content and terminate the recursion. This is O(n2^n) time and O(n) space, where n is the buffer length. If stack overflow becomes a concern, consider rewriting this using iteration and a stack to explicitly manage the call stack yourself.

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