Given that there are 12 questions to be answered in an examination, find the number of ways of answering 8 questions correctly?
I'm thinking permutation because order of the questions is important.
so, 12!/(12-8)!
correct
that way i used for calculating combinations when betting
play let's say in your case 8/12 meaning that i need do hit 8 correct and can miss 4 and still get the greens
and that way u get how much there are combinations without repetition same combination
If there were 2 questions,
you get both right,
so you get the first and second questions right.
Does it matter which question you answered first ? no
There are 2 ways you can get both right depending on which question you answer first
but, as such, there is a single "selection or combination" of getting 2 answers right,
there are 2 "arrangements or permutations" of getting 2 answers right.
You want 8 correct answers.
The different ways that can happen depends on the order in which you answer the questions.
You could start at Q1 or Q2 etc...
that would be permutations.
Maybe the question assumes you answered the questions in the order 1, 2, 3, 4, 5, 6, 7...12
Then it's just a matter of selecting (combinations) 8 from 12.
However, even if you chose your own order,
and you got 8 correct,
that's a certain group of 8 questions from the 12.
Does it really matter in which order you answered them in ??
Sometimes you just have to make assumptions based on context. For example, people are generally considered non-identical, but objects like balls that have the same colour and are not numbered would be considered identical. A meticulously worded question will often specify "where order is not considered" or some such.