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How many ways to select $4$ homework out of $6$ in a specific order?
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?
- combinatorics
- permutations
2 Answers 2
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 $6 \cdot 12=72$ .
- $\begingroup$ Okay. So you first found out the possible ways she can finish her last 2 homework sets from a sample space of 4. Then what is left is the first 4 homework subjects where you used permutation because the order of 2 homework sets are important therefore 2P4 * 2C4 = 12*6 = 72. I haven't fully comprehended how you figured it out just based on the question alone but I will try and solve more counting problems this way. $\endgroup$ – swordlordswamplord Commented Feb 11, 2020 at 6:13
- 1 $\begingroup$ I did not do 2P4, but (4P4)/2. They are numerically equal, but not the same approach. It is not that I chose one class for one time slot and another for a second and quit. I said half of all permutations of the 4 are acceptable. $\endgroup$ – Ross Millikan Commented Feb 11, 2020 at 6:38
- $\begingroup$ So would it be incorrect to do 2P4 * 2C4? $\endgroup$ – swordlordswamplord Commented Feb 11, 2020 at 6:42
- $\begingroup$ Yes absolutely wrong , but again somehow gives same answer ! $\endgroup$ – aryan bansal Commented Feb 11, 2020 at 6:46
Minnie can choose which two of the other four subjects to study in $\binom{4}{2}$ ways. She now has four slots to fill in her study schedule. She can choose two of them in which to study calculus and statistics in $\binom{4}{2}$ ways. Those two slots can be filled in only one way since she studies statistics before she studies calculus. She can fill the remaining two slots in her schedule with the remaining two subjects in $2!$ ways. Hence, there are $$\binom{4}{2}\binom{4}{2}2!$$ ways for Minnie to select four subjects, including both statistics and calculus, and arrange them so that she studies statistics before calculus.
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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 ...