![]() ![]() If you really want to start with the denominator (which does not make much sense here), and after the numerator 25! has been calculated afterwards, simply press. Instead of first dividing "the wrong way round" and then correcting this by you can simply use. (08-17-2017 12:25 PM)Gamo Wrote: and I have no choice to reverse the division by using 1/x since the problem start from the denominator portion of the fraction. (08-17-2017 12:25 PM)Gamo Wrote: I decided to calculate manually this way because of the rule of operation that must start from the inner most parentheses firstįirst of all, you do not have to start at the innermost parentheses. With this method all intermediate results are integers so that roundoff errors are minimized. Take a look at the linked thread and you get an impression of how it can (and should) be done: 25 / 1 7 and then determining the permutations from these three.This means that C(25, 7) is almost certainly not calculated by first evaluating 1 We can safely assume that it does not call the factorial routine. Internally the 15C probably calculates permutations and combinations by repeated multiplication, similar to the method in the thread I linked to. ![]() For instance, calculating 1/3 * 3 with the usual 10 digits yields 0,99999 99999 while with 13 digits you get 0,99999 99999 999 which, after rounding to 10 display digits, shows up as "1". On the other hand user programs like the ones we're discussing here always work with only 10 digits. This usually (but not always) gives results with at least 10 exact digits – which is what you see in the display. So the 15C's internal nCr and nPr functions are performed with 13 digits. Quote:That's why most HPs, including the 12C, internally use 3 more digits for their calculationsīoth the 12C and the 15C internally (!) use 13 digits so that the final 10-digit result that is presented to the user is correct (ok, mostly). I though that HP 12C and HP 15C is the same family and should be limited to the same 10 digits calculation limit or the HP 15C use difference formula to get the result? (08-17-2017 10:01 AM)Gamo Wrote: I try the same 25P7 on the HP 15C build-in functions and the result is But all this is quite cumbersome on the 12C which does not feature any loop counters.Įdit: anyway, here is a short permutation routine that shows how it can be done. no overflows unless the result overflows. Doing this without factorials also allows larger arguments, i.e. ![]() this thread, especially Werner's post #4. )įor more information on this topic and ways to calculate permutations and combinations with minimized roundoff errors cf. ![]() 25 which is faster and causes less roundoff problems.OK, actually these calculators will evaluate permutations by 19 which rounds to 2422728000 in the returned 10-digit result. This will happen with any 10-digit calculator if the result is calculated by dividing factorials. The numerator is slightly low (rounded down) and the denominator is slightly high (rounded up) which causes a slightly low result. What you see is the result of limited precision and roundoff. (08-17-2017 05:16 AM)Gamo Wrote: I'm not sure if this result is more accurate or others just round the result up. The returned result is 2 442 727 999 and it is not exact. Most P answer correctly but with this 25P7 I check most calculator result to 2,422,728,000 but on this program result to 2,442,727999 Represents the number of ways of selecting $k$ objects from a set of $n$ objects when repetition is permitted.Įxample.(08-17-2017 05:16 AM)Gamo Wrote: I try your Permutation and Combination program and found one little problem with the Permutation answer. In this case, we are selecting the subset of $k$ boxes which will be filled with an object. Since the order in which the members of the committee are selected does not matter, the number of such committees is the number of subsets of five people that can be selected from the group of twelve people, which isĪlso counts the number of ways $k$ indistinguishable objects may be placed in $n$ distinct boxes if we are restricted to placing one object in each box. In how many ways can a committee of five people be selected from a group of twelve people? Is the number of ways of selecting a subset of $k$ objects from a set of $n$ objects, that is, the number of ways of making an unordered selection of $k$ objects from a set of $n$ objects.Įxample. ![]()
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