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    <p>You're looking for the <a href="http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29" rel="nofollow">Stars and Bars</a> formula. Positions of array A are the bins; into each position the corresponding element from A and any number of elements from B will be inserted. (Which elements are chosen is already determined by their original order, so we treat them as indistinguishable.) If there are (n-1) elements in A (add a bin to one end) and k elements in B, the formula gives is the binomial coefficient</p> <pre><code> n + k - 1 ( ) k = ( (n+k-1)! / ( k! (n-1)! ) = (a.size() + b.size())! / (a.size()! * b.size()!) </code></pre> <p>which is the same as templatetypedef's answer :)</p> <p>Actually <em>calculating</em> the quotient of such large numbers is another issue. In most languages, the numerator will commonly produce integer overflow. One simple strategy, not necessarily optimal, in C++ would be</p> <pre><code>unsigned long long gcd( unsigned long long &amp;a, unsigned long long &amp;b ) { return b? gcd( b, a % b ) : a; } std::vector&lt; std::size_t &gt; numerator( a.size() ); // factors of (a+b)!/b! std::iota( numerator.begin(), numerator.end(), b.size()+1 ); for ( std::size_t afactor = 2; afactor != a.size()+1; ++ afactor ) { std::size_t reduced = afactor; for ( auto &amp;&amp;nfactor : numerator ) { auto common = gcd( afactor, nfactor ); nfactor /= common; reduced /= common; if ( reduced == 1 ) goto next_afactor; } throw std::logic_error( "Fractional combinations" ); next_afactor: ; } return std::accumulate( numerator.begin(), numerator.end(), 1, std::multiplies&lt; std::size_t &gt;() ); </code></pre>
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