Combinations and Permutations and Factorials confusion

[orangepopsicle]

Hey guys,


As I'm solving these factorial/combination/permutation problems I am having a huge problem differentiating when to use what formula and what counts as a problem where order matters and when it doesn't, I'm so confused can someone help me?

So I know that when order matters and there is no repitition, the equation is 👎!/(n-r)!
When order matters and there is repetition allows the equation is n^r
when order doesn't matter and there is no repetition the equation is n!/r!(n-r)!

So how would you go about solving these two problems?

In how many ways can 4 of the same physics book, 3 of the same algebra book, and 2 of the same chemistry book be arranged on the shelf? They are telling me since there is repeats and since order matters, you use the formula n!/(a!b!c!)
where a.b and c are the number of repeats of a specific book. How does this tie in to the combination permutation equations above, and why does order matter in this? Doesn't it not matter what book we choose first, second, third in the order on the bookshelf as long as we can find all the different ways it can happen?

Similiarly, This second problem that states that there are 5 chairs and that there are 2 girls and 3 boys, and to find all the different ways the boys and girls can sit in the 5 chairs as long as the 2 girls are always sitting next to each other. Isn't this a problem where order matters? since the 2 girls need to always be sitting next to each other. And which equation would you use from the 3 listed above? I am so confused. someone help!

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[510586]

I thought there were only 2 equations. When order doesn't matter it's n ! / k! (n-k)! and when it does it's the same equation without multiplying by k! ?

 

[orangepopsicle]

Yea i know about those two equations, but neither of those two problems listed above use those equations at all. You also need to take into consideration whether there are repeats or not.

 

[510586]

Yea i know about those two equations, but neither of those two problems listed above use those equations at all. You also need to take into consideration whether there are repeats or not.

oh ok. Hopefully someone can give a systematic way of doing these. I usually just draw the chairs out or something if it's only 5 people.

 

[skedaddle]

Chad had a great explanation of this...a specific video called permutations and combinations. If you need, later I can explain. If you don't have Chad 🙂

 

[alexis2010]

This is how I usually do these type of problems. For the first problem think of it as an arrangement not a combination or a permutation. You're arranging things in a line. The formula for that is simply n! Since we have 9 books it's 9! But because we have objects that are identical we divide by how many times each object repeats. Here's why we divide. If I arrange 4 identical books in a row and I switch the first one with the last one we can't tell the difference.
For the second problem it's also an arrangement.
I usually think of a problem as a combination or a permutation when the question asks to select r things out of n things. For example how many ways you can choose 4 students out of 10 to participate in a competition? That's a combination problem because the order doesn't matter. If we choose john Jimmy Alex and Maria it's the same as if we choose Maria Alex Jimmy and john.
For a permutation the order matters.
I'll explain more if you still need help.
Gotta go my break is over.

 

[orangepopsicle]

Hey skedaddle, I would really appreciate if you could explain it to me. My subscription for Chad's just ran out and i am no longer able to access his videos.

 

[orangepopsicle]

This is how I usually do these type of problems. For the first problem think of it as an arrangement not a combination or a permutation. You're arranging things in a line. The formula for that is simply n! Since we have 9 books it's 9! But because we have objects that are identical we divide by how many times each object repeats. Here's why we divide. If I arrange 4 identical books in a row and I switch the first one with the last one we can't tell the difference.
For the second problem it's also an arrangement.
I usually think of a problem as a combination or a permutation when the question asks to select r things out of n things. For example how many ways you can choose 4 students out of 10 to participate in a competition? That's a combination problem because the order doesn't matter. If we choose john Jimmy Alex and Maria it's the same as if we choose Maria Alex Jimmy and john.
For a permutation the order matters.
I'll explain more if you still need help.
Gotta go my break is over.

I guess I'm having trouble figuring out when order does and does not matter. Like for example, when order matters does that mean that the order in which you pick someone for a specific thing matters? I get why picking a combination for a password or a lock the order would matter, but i am still confused because for a problem they say that picking someone for 1st, 2nd or 3rd place trophies, order does not matter, and this is something i thought that order would matter, since it matters who you pick for 1st and 2nd place as opposed to 3rd place since if order did not matter, who gets 1st 2nd or third place trophies wouldn't matter either.

 

[alexis2010]

You're right. When picking someone for 1st, 2nd and 3rd order does matter. Let's say john was first Jimmy was second and Mike was third. It's different if Mike was first john second and Jimmy third. In this case it's a permutation. Usually the wording of the question should give you a hint if it's a combination or a permutation.

 

[skedaddle]

Let me go over the vid one more time tonight and I'll get back to you late tonight. With formulas and how to use them. But the easy way to memorize it is this: if combination/order matters, it's going to be a higher number (think how many possible combinations are on a safe dial). If order doesn't matter, it's going to be much smaller because there are less permutations.

 

[skedaddle]

Okay I'll just do it now with notes. So starting with the general principle first: If order matters, DON'T divide, just mulitply. If order does NOT matter, divide by the # of selections made. This makes sense b/c if order matters (think that safe example), it's going to be a large # so just multiply.

Okay, the long way around: the formula for permutations problems is nPk= n!/(n-k)! The formula for combos is nCk=n!/k!(n-k)!

Let's use an example. You're taking 3 coworkers to a ball game but have 6 coworkers. Does order matter? Think...think...think...no. It won't matter what 3. Let's use the formula then the shortcut. Since order doesn't matter we're using nCk. So n factorial is 6!, or 6x5x4x3x2x1. k factorial is number of selections factorial, we are selecting 3 so 3x2x1. (n-k)! is (6-3)!=3! so 3x2x1. Put it all together. 6x5x4x3x2x1/(3x2x1)3x2x1. Now you should take notes: We have similar #'s in the numerator and the denominator. After cancelling we're left with 6x5x4/3x2x1.

So the short way: We have 6 choices, we're choosing 3. Since we're choosing 3 people it'll be 6x5x4 as you take the 3 highest numbers, then divide by # of selections factorial. So 6x5x4/3x2x1. (n-k)! cancels out the top anway, so just leave it out and leave the factorial off the top. You will save yourself large amounts of time on problems like these. Remember, for permutations problems where order matters, just multiply as the equation is different, and the last #'s in the numerator will cancel with all the #'s in the denominator. I really hope this helps! If not, read it one more time and do a few practice examples. Mine is similiar to Chad's but I made it up. I urge you to run both ways through a calc and compare 🙂

 

[skedaddle]

Order matters v order does not matter:

Order matters in a safe dial. When arranging books on a shelf, when choosing people for scholarships when the scholarships differ by $ awards (like 1st scholarship is 25k, 2nd is 15k, and 3rd is 5k). Who you pick 1st vs 3rd will matter. So that's total permutations. Order does not matter when picking random events, friends to attend a movie, picking 4/10 books to randomly read...things like that.

Here's Ari on the books on the shelf question:

"You have 9 books total right? And presumably 9 spots on the book shelf. If all of the books were different, we'd just do 9 factorial. But since we have replicas of the same books, we're going to get duplicate combinations. To account for duplicates items in a set, we divide by the factorial of the number of duplicates. For example, 4 duplicate physics books means we divide by 4 factorial, and 3 duplicate algebra books means we divide by 3 factorial. When all is said and done, you'll have 9! / (4!*3!*2!). 9 for the original amOunt of books, and divide by the factorials of the duplicates to account for the duplicate sets from ordering them on the book shelf."

 

[orangepopsicle]

thanks so much, this really cleared it up for me.