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    <p>This should do the trick:</p> <p>The shuffle gives some form of randomness to which boy and girl are paired with each other</p> <pre><code>In [115]: boys = ["Boy1", "Boy2", "Boy3", "Boy4", "Boy5"] In [116]: girls = ["Girl1", "Girl2", "Girl3", "Girl4", "Girl5"] In [117]: random.shuffle(girls) In [118]: girls Out[118]: ['Girl5', 'Girl4', 'Girl3', 'Girl1', 'Girl2'] In [119]: for i in itertools.izip(boys, girls): .....: print i .....: ('Boy1', 'Girl5') ('Boy2', 'Girl4') ('Boy3', 'Girl3') ('Boy4', 'Girl1') ('Boy5', 'Girl2') </code></pre> <p><strong>EDIT</strong>: If you want every possible pairing, check this out:</p> <pre><code>In [126]: boys Out[126]: ['Boy1', 'Boy2', 'Boy3', 'Boy4', 'Boy5'] In [127]: girls Out[127]: ['Girl1', 'Girl2', 'Girl3', 'Girl4', 'Girl5'] In [128]: [girls[i:]+girls[:i] for i in xrange(len(girls))] Out[128]: [['Girl1', 'Girl2', 'Girl3', 'Girl4', 'Girl5'], ['Girl2', 'Girl3', 'Girl4', 'Girl5', 'Girl1'], ['Girl3', 'Girl4', 'Girl5', 'Girl1', 'Girl2'], ['Girl4', 'Girl5', 'Girl1', 'Girl2', 'Girl3'], ['Girl5', 'Girl1', 'Girl2', 'Girl3', 'Girl4']] In [129]: for combo in (itertools.izip(boys, g) for g in ( girls[i:]+girls[:i] for i in xrange(len(girls)) )): .....: for pair in combo: .....: print pair, .....: print '' .....: ('Boy1', 'Girl1') ('Boy2', 'Girl2') ('Boy3', 'Girl3') ('Boy4', 'Girl4') ('Boy5', 'Girl5') ('Boy1', 'Girl2') ('Boy2', 'Girl3') ('Boy3', 'Girl4') ('Boy4', 'Girl5') ('Boy5', 'Girl1') ('Boy1', 'Girl3') ('Boy2', 'Girl4') ('Boy3', 'Girl5') ('Boy4', 'Girl1') ('Boy5', 'Girl2') ('Boy1', 'Girl4') ('Boy2', 'Girl5') ('Boy3', 'Girl1') ('Boy4', 'Girl2') ('Boy5', 'Girl3') ('Boy1', 'Girl5') ('Boy2', 'Girl1') ('Boy3', 'Girl2') ('Boy4', 'Girl3') ('Boy5', 'Girl4') </code></pre> <p><strong>EDIT 2</strong> (fixes EDIT 1):</p> <pre><code>&gt;&gt;&gt; perms = itertools.permutations(girls) &gt;&gt;&gt; len([tuple(p) for p in (itertools.product(boys, g) for g in perms)]) 120 &gt;&gt;&gt; perms = itertools.permutations(girls) &gt;&gt;&gt; len(set(tuple(p) for p in (itertools.product(boys, g) for g in perms))) 120 </code></pre> <p>I had to go with <code>len</code> because there are 120 possible pairings and I didn't want to clutter the post. This is why there is <code>len(...)</code> and <code>len(set(...))</code></p>
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    1. POPairing people in a contest
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    1. COIn fact I wanted all possible combinations of pairs so firstly I have to use permutations of girls, not just one shuffle. Probably I didn't make my point clear. But that helped me anyway, thanks.
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    2. CO@ecik: check out my edit
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    3. COaren't you suppoesed to make permutations of `girls`? For example, your method doesn't yield a combination: `(B1G1,B2G3,B3G2,B4G5,B5G4)`.
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