Minnie has homework from $6$ subjects including calculus and statistics this week. Among them, she wants to finish homework for $4$ subjects including calculus and statistics today. She wants to finish statistics before calculus. How many ways are there for her to select the $4$ subjects, and then decide the order of doing homework?
So I used a very roundabout method which is not ideal since I would need to apply the theory when I take the exam.
So basically, I simplified the subjects into $5$ choices where I merged statistics and calculus into one unit where it is implied that statistics will be finished before calculus.
Then, I used a tree diagram that basically starts with statistics-calculus as the first homework set completed, which branches out into $4$ next case scenarios (homework $2$, $3$, and $4$) whereby, from each of those branches, it further expands until all the homework order is exhausted.
For the case when statistics-calculus is the first set of homeworks, completed, there are $24$ possible outcomes. Then when statistics-calculus is the second and third set of homeworks completed, it comes out to $48$. Since I am only interested when statistics-calculus is either in position $1$, $2$, or $3$, I just added $24+48 = 72$ possible outcomes as my answer.
However I tried every possible way to use either permutation or combination to arrive at $72$ and just couldn't find it. Can someone explain to me please how to approach this problem methodically?
She can choose the two other subjects in ${4 \choose 2}=6$ ways. Having chosen the subjects, she could put them in $4!=24$ orders, but half of them have Statistics and Calc out of order, leaving $12$. Then $6 \cdot 12=72$.