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Factorials: When to Divide by Another Factorial versus NOT?

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Factorials: When to Divide by Another Factorial versus NOT?

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Hi there! I'm getting really confused on when to divide a factorial by TWO factorials, versus when to only divide by ONE factorial. I've been following the MGMAT books and understand the Anagram analogy, but when I apply the principles to "real life" (aka, practice problems), my understanding doesn't always match up with what the answers actually are. I thought that we only had divide when there was a set of "repeats" in the factorial. Case in point: two questions below. Why do I divide one truffle box-making example by another factorial, but do NOT do this with the passengers of the plane?

1. Truffles:
"Amy and Adam are making boxes of truffles. They have an unlimited supply of 5 types. Each box holds 2 truffles of different types. How many boxes can they make"?

I thought the answer was 5!/3! (I chose 3 because there are 3 "no's", or truffles that don't make it into the box).

BUT, the answer in the book is: 5!/(2!*3!) = 10. I understand that this is because it's "2 yesses" (2 truffles made it into the box) and "3 no's" (3 truffles did not) but since the truffles are of different types, couldn't I technically categorize this as "1-2-NNN" (N = no) meaning it would just be 5!/3! ? My reasoning for that is supported (at least, to me) by the following plane passenger question:

2. Plane Passengers:
"If 7 people board an airplane with only 3 available seats, how many different seating arrangements are possible?"
The answer here is 7!/4! (aka, 1-2-3-NNNN) (N = no seat for person). Why is it this, and NOT "7!/(3!*4!)

....

I hope that makes sense - would definitely appreciate any explanations, rules to follow, examples to keep in mind, etc! Thank you!

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The Truffles problem is a combination problem, and the Plane problem is a permutation one.

1. Truffles

Think of it this way: You have 5 types. To be able to make a box, you need to choose 2 different types. This is: 5C2 = 5!/2!*(5-2)! = 5!/2!3!

Alternatively, let's say you need to fill two positions in the box. For the first place, you can select any of the 5 types. For the second place, you can select any of the remaining 4 types (to make a box, you need different types. So you rule out the type of the truffle you picked for the first place). This makes it 5 * 4 = 20 ways of filling those places.

Now, we also know that if (truffle1, truffle2) is a box, then (truffle2, truffle1) is also the same box. How many ways can we arrange truffle1, truffle2? 2 ways (= 2!). So, in your original answer (20), you are double counting (by counting the above two pairs as different, when infact both are same box).

So, divide 20 by 2! = 10, which is also 5!/2!3!

2. Plane

There are 7 people, 3 places to be filled.

First seat can be filled 7 ways.
Second seat can be filled 6 ways.
Third seat can be filled 5 ways.

=> Answer is 7 * 6 * 5 = 210. This is also the same as 7!/(7-3)! = 7!/4!

If I have to approach this your way, I'd do as follows:

First person can go to any of the 3 seats.
Second person can go to any of the remaining 2 seats.
Third person can go to the remaining 1 seat.

So, three people I pick can occupy their places in 3*2*1 = 6 ways.

But, how do I select these 3 people from 7? 7C3 = 7!/3!4!

Therefore, the total number of ways I can do this is 7C3 * 6 = 210

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Awesome explanation - thank you so much! Much appreciated!

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Hey All,

I'd like to suggest what I think is a better overall method for combination/permutations than the anagram grid. It's called the slots method, and it works like this:

1. Find # of slots (decision points)
2. Fill in each slot with number of choices at that decision point
3. Work out if Order Matter or Not (If vanilla-chocolate isn't the same as chocolate-vanilla, order matters)
4. If Order Matters, multiply all slots (don't divide by anything); if Order Doesn't Matter, multiply all slots and divide by the number of slots factorial

Let's try with your questions:

Truffles

1. 2 slots
2. 5 choices for first slot, 4 for the next
3. Order doesn't matter
4. (5 * 4)/2! = 10

Plane Passengers

1. 3 slots
2. 7 choices for first slot, 6 choices for second, 5 choices for third
3. Order matters (arrangements)
4. (7 * 6 * 5) = 210

Hope that helps!

-t

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Hi Tommy,

Thank you so much! I really appreciate your explanation of the slot method (I've been studying it, but wasn't clear on how to use it effectively in lieu of factorials). So, this not only tackles the question of, divide by a factorial at ALL, but also if I divide by one OR MORE factorials - correct?

Thanks again! This forum is such a lifesaver!

Amy

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Hey Amy,

That's the great thing about the slots method; it NEVER changes. The only weirdness comes when order only matters in certain places. Here's an example. See if you can solve it with the slots method (and other instructors/experts, please hold off on writing up the explanation). I'll step in after you've given it a try:

A soccer team is to be put together from a pool of 9 kids. Only two of the kids can play the goalie position, and they can play no other position. The team also needs 2 forwards, 2 midfielders, and 1 defender. How many different teams can be made? (Assume that if the same kid is moved to another position, that would count as a different team)?

-t

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Hi Tommy,

Thank you for the challenge! I'm nervous but this is really helpful. So, here is my thought process:

Order matters (since we're 'arranging' a soccer team), so:

Goalies = 9*8 = 72
Forwarders = 7*6 = 42
Midfielders = 5*4 = 20
Defender = 3*2 = 6

= 72+42+20+6 = 170 ways of arranging a soccer team of 7 kids out of 9, total.

----
Is that correct? My gut is telling me that there must be another step here, because we are trying to find the number of different teams that we can make, and not how many ways we can arrange for multiple teams (and, you told me I should expect something 'weird'). So:

Since we're choosing 7 kids out of 9 kids for each team, do we then have to do 9!/7! = 72, and then multiply this by the 170 above, to get 170*72 = 13240 (?) That seems like WAY too many ways to arrange 7 out of 9 kids into 4 different available positions in each single team, though!

----

I understood your reasoning for the earlier ones, but I'm still (admittedly) confused when we move 'beyond the basics.' I also seem to have a hard time distinguishing between what I think means "order matters" versus what the GMAT determines is an "order matters" - so beyond your explanation for the challenge question you posed, any tips on how to best distinguish between the two would be GREATLY appreciated!

Thank you so much!

Amy

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...one other thought was that this would be 9!/(2!)(2!)(2!) = __, but that number was even higher..

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Hey Guys,

Good talk and try, but there are a couple problems here. Let's use the method:

1) There are 6 slots (Goalie, Forward, Forward, Midfield, Midfield, Defense) -- I fooled you on the goalie thing. There is only one goalie (it says "Only 2 kids can play the goalie position" -- the word "the" implies there's just one goalie).

2) There are 2 choices for goalie. 7 for first forward, 6 for second forward. 5 for first midfield. 4 for second midfield. 3 for defense.

3) Order does matter between the different positions. Unfortunately order does not matter within the positions!

4) 2 * (7*6)/2! * (5*4)/2! * 3

2 * 21 * 10 * 3 = 1260

Does that make sense?

-t

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Ahh, I gotcha. That makes sense. The one hang-up I still have for this problem is why we didn't start with 9!/7! for the first one, the goalie (nine choose 2) - is that because it's just ONE goalie position that is available?

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(ps: thank you!)

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Hey Ostrow,

Two answers:

#1: yes, you're right. We aren't choosing two; were choosing one. So it wouldn't make too much sense to say nine choose two!

#2: with the slots method, you don't need to think in terms of blah choose blah. That's the joy!

You're welcome!

-t

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