My english teacher gave us 8 possible essay questions that might show up on are exam. She said she will put three of those essays on the test for us to do. She said that we could figure out the ones that she was going to put on there. If there is 8 possible questions what numbers might be on the test.
I must be missing something, since there will be 56 different triplets one can from from 8 questions.
8x7x6/(3x2x1) = 56
or
C(8,3) = 56
To figure out the possible combinations of questions that might appear on the test, you can use a combination formula.
In this case, we have 8 possible questions and need to choose 3 questions for the test. The formula for combination is:
C(n, r) = n! / (r!(n-r)!)
Where n is the total number of items and r is the number of items chosen.
So, plugging in the values, we have:
C(8, 3) = 8! / (3!(8-3)!)
Simplifying this equation:
C(8, 3) = (8 * 7 * 6) / (3 * 2 * 1)
= 56
Therefore, there are 56 different combinations of 3 questions that can appear on the test.