115 heroes. 5 heroes to Radiant, 5 different heroes to Dire.
There is 115 Choose 5 = 115! / (5! * 110!) ways to choose Radiant's heroes.
Given that Radiant has already chosen, there is now 110 remaining heroes. There is 110 Choose 5 = 110! / (5! * 105!) ways to choose Dire's heroes.
Thus, the answer is to multiply the number of ways to choose Radiant's heroes with the number of ways to choose Dire's heroes. That is: 115 * 114 * 113 *... *106 / (120)^2 = 18784179344417256
Just had this idea while permutation topic was going in class , but havent completely studied it so i understood first part and not the second one , but dont worry i will later . Sixe ty
^^Its a thing relating to factorials and permutation , I wouldnt bother researching it if not good at maths or no interest in maths . Just know that u can play 18784179344417256 matches without repetition in the draft of the overall game . It is a mindblowing amount , and another proof that dota is something that even if u play hundreds of thousands of hour ,u cant just get so good at it ,that u dont have to think anymore .
Reminds me of the permutations and combination classes at school n college. Good old times , one of the most interesting topics in math which you can apply in all industries.
i dont think its right.
Had the same idea as sixe but there will be a lot of match ups that appear double.
You can not just multiply (115 5)*(110 5)
Example:
5 heros in total and you can pick 2 per team
Answer would be 30 with math above
its actually only 15
you can write it down
Or am I retarded its late
^^^^
U guys just dont know permutation and combination . I cannot tell u the reasons , it will be like teaching algebra to someone who barely knows counting .
Lol what. I just pointed out a mistake. I am right.
I give you the correct answer when i have time on the weekend and find my old math book
@dukeus : Its 30 because you count radiant and dire different
If you take radiant and dire as the same e.g: AB vs CD is the same as CD vs AB then u need divide the formula by 2!
i dont wanna be some r/iamverysmart material here but do you guys actually struggle with this entry level combination problem and fail to understand it?
@duchass: 115C5 x 110C5 is same as 110C5 x 115C5.
We r doing 10 selections only. Counting one as team 1 and the other team 2 with 5 players each.
There's no need to divide again
@LucaManny
Yeah thats what i meant.
I thought ab vs cd is the same as cd vs ab. Its the same match up. Thats what i try to say.
So you have to divide the result by 2 to get the answer op wanted in my opinion.
Sry should take my time and explain things better next time i write something. Thx for help
@perdita , u are trying to be r/iamverysmart cuz i just told i havent studied the whole topic .
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If someone has good Maths here , i would like to ask him the following question which came to my mind but couldn't solve .
There are 115 heroes , 5 in our team , 5 in enemy team . If we play some amount of games , such that in every game , at least one hero is different in our team OR one hero is different in their team , how many total drafts are possible ?
Also explain which method u used as i couldnt solve .