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How do I calculate permutations from combinations?
If you already have a combination and want to turn it into a permutation, you need to impose order on the set of items, i.e., choose one of the possible orderings for your set. Hence, the number of permutations of r items chosen from n items is equal to the number of combinations of r items chosen from n items multiplied by the number of orderings of these r items, i.e., by r!.
How do I calculate combinations from permutations?
If you already have a permutation and want to turn it into a combination, you need to remove order, i.e., regard all possible reorderings as the same object. Hence, the number of combinations of r items chosen from n items is equal to the number of permutations of r items chosen from n items divided by the number of orderings of these r items, i.e., by r!.
What is the formulas for combinations and permutations?
for questions as such, id have to use the C (combination button} multiple times and since the option in 991cw is in catalog>probablity>then symbol, it consumes alot of time especially in exams with a strict timing, like going into the menus numerous times.
so i wanted to ask, is there a way i can move factorial and combination symbol to any of the variables, change the keys fuctions, or any tips? my exam is soon any help will be apreciated thank you
In this case I would start to simplify the equation. In your example you're looking for 500 choose 2, which is 500!/498!/2!. This can be easily changed to 500*499/2, which can be calculated.
In general terms if you have n choose k, you only need to calculate a "partial factorial" from n to max(k, n-k) and then divide by min(k, n-k)! due to the results being mirrored. This makes the calculation much easier.
Also in certain cases you could start dividing with the min(k, n-k)! while multiplying, but that will lead to remainders etc.
Use the Pascal's triangle property:
C(n,k) = C(n - 1, k) + C(n - 1, k - 1) and dynamic programming. No factorials involved.
The triangle of Pascal being:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1