![]() ![]() ![]() In general, n equals the product of all numbers up to n. The exclamation mark () represents a factorial. To calculate the number of combinations with repetitions, use the following equation: Where: n the number of options. Similar procedure 'moves' coefficients to non-negative. Permutations: The order of outcomes matters. You can simplify the permutation equation by substitution where you would come up with this equation: P(n,r) C(n,r) r When to use either the combination or permutation formula will depend on whether you have to take into consideration order or not. To 'move' solutions to non-negative it is needed to add to all solutions value -min(solutions) and to result -min(solutions)*sum(coefficients). The two equations for permutations and combinations almost look similar. Permutation(n1 and r3Permutation(n2 and r2)Permutation(n3r1) Since the problem didnt specify which genre will come first or second, we dont know which of n1,n2, or n3 will get r3 or 2 or 1, so I understand this way isnt correct, but Im struggling to understand when to use Permutation rule and when to use simply just multiplication. Second is that it is always possible to transform problem to problem of the same type with non-negative coefficients and solutions. ![]() It is enough to see that for four numbers 0 = a1*b2 + a2*b1. Min = sum(c*s for c, s in zip(coeffs, reversed(sols))) As such, the equation for calculating permutations removes the rest of the elements, 9 × 8 × 7 ×. max = sum(c*s for c, s in zip(coeffs, sols)) First, if coefficients and solutions are non-negative numbers and if they are represented as a sorted list, then maximal value equation can have is 'parallel sum', and minimal value is 'reversed sum'. This is brute-force solution with pruning search where solution can't be found. ![]()
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