@Leaderboard
I’m just lost on that problem while doing with combination method. So i dont think i can put what i tried and what i am confused about. I just dont know how to approach it.
Tell that then. To answer your question, let the words in the first randomised trial be P, Q and R respectively. What is the number of ways possible for the first randomised trial?
7!
Why? As a hint: you have 10 ways to choose P.
Are you asking for the sample space or the not same case ?
If sample space then 10!,
if not same i think 7! ?
The first case. And why 10!? We are not picking 10 elements, but 3.
Aren’t 10 words listed ?
1st position has 10 posibilities, 2nd has 9 possibilities and so on ?
I dont understand.
Correct in that there are 10 words originally, but you are picking three from the 10.
But why ?
Are we doing the 1 - P(satiesfies) rule ?
Because as far as i understand the question is asking for the case that it dont match
No. I recommend reviewing the lessons on Combinatorics and Permutations on PrepSwift (including doing the associated quizzes) before going further, as it does sound like you have a foundational gap.
Ok will do. But can you tell how you would do this with combinations ?
I’m hoping that you get the answer as you revisit the concepts. If not, I’ll explain then.
If anyone is watching, please help. I dont understand what this guy is asking.
If the first randomization gives you this tuple:
(n_1, n_2, n_3, n_4, n_5, n_6, n_7, n_8, n_9,n_{10})
then how many arrangements could there be for the second randomization under no restrictions/constraints?