[Key Techniques] Discuss Permutations II

Main post link -> http://blog.brilliant.org/2013/01/09/permutations-ii/

  1. Samir and Mirdula each have distinct 7 books that they want to display on their shelves. Mirdula has lots of room on her shelf, however, Samir only has room to put 6 of his 7 books. Assuming that they fill up the entire shelf, who has more possible ways to arrange the books on their shelf? Why is this the case?

  2. A waiter at a restaurant takes the drink order for a table of ten people but forgets to write down who ordered which drink. The orders are 5 cups of coffee, 3 glasses of water, 1 glass of milk, and 1 apple juice. How many different ways can the waiter deliver the drinks to the customers so that each customer gets one drink?

  3. Consider all the ways to rearrange the letters of the word CALVIN, and arrange them in alphabetical order. In what position will CALVIN appear in? How about NIVLAC?

  4. (*) Lisa wants to arrange 7 ornaments on her mantle. She has 2 identical cats, one mouse and 4 other different ornaments. If the mouse cannot be next to either of the cats, how many ways can they be arranged? Hint: Principle of Inclusion and Exclusion.

Note by Peter Taylor
8 years, 5 months ago

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Comments

On the blog, Saurabh Dubey posted that:

1->Samir has more options because he has to select 6 books out of seven (7!) and then arrange them so total of 6!*7! whereas Mirdula only has to arrange 7 books (7!).

2->The number of ways would be 10!/(5!3!1!*1!) i.e 5040 ways.

Discuss.

Calvin Lin Staff - 8 years, 5 months ago

From Facebook:

Alexander wrote: "Consider the case where Samir only had one slot, it is obvious he has a smaller amount of ways to arrange the books than if he had more slots. The the most slots someone has the more ways they can arrange the books. From this logic Mirdula has more ways to arrange the books because she of the above logic and the fact she does not have any restrictions."

Discuss.

Calvin Lin Staff - 8 years, 5 months ago

There is a possible ambiguity in the first. Do they always try to fill up the whole space?

Also, #3 is very similar to project euler problem 24 Its one of the few that can be done by hand

Harshit Kapur - 8 years, 5 months ago

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That is true, I've added that they fill up the entire shelf.

Lexicographic / alphabetical ordering is a common problem. You should think about what the general solution is like, and what is a quick way to approach it.

Calvin Lin Staff - 8 years, 5 months ago

1.Samir and Mirdula has equal number of ways to arrange the books. :) 7 ⁣P7=7!=7×6×5×4×3×2×1=7×6×5×4×3×2=7 ⁣P6 {^7\!P_7}=7!=7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 = {^7\!P_6} .

Soham Chanda - 8 years, 5 months ago

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OMG what happened to latex?? Calvin help me!!

Soham Chanda - 8 years, 5 months ago

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You have to put the latex code in < \ ( code \ ) > [with spaces removed between \ and ( ,and also \ and ) ]otherwise it will not know that you need to display the math.

I've edited the post for you, so you may click edit to see what I did.

Calvin Lin Staff - 8 years, 5 months ago

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@Calvin Lin THANKS :)

Soham Chanda - 8 years, 5 months ago

Can you think of an explanation which shows that these 2 values are equal, without knowing the actual calculation?

Calvin Lin Staff - 8 years, 5 months ago

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Let the books of Samir be named a,b,c,d,e,f,ga,b,c,d,e,f,g Let one of his arrangement be a.c.b.d.g.ea.c.b.d.g.e Then he will have only one book left ie. ff If he had space for another book he would be able to make only one arrangement ie. a.c.b.d.g.e.fa.c.b.d.g.e.f which will correspond to one of Mirdula's arrangement.Similarly every arrangement of Samir's book will have a similar counterpart in Mirdula's arrangement of books.

Soham Chanda - 8 years, 5 months ago

3.CALVIN comes in 131^th place. First I saw the number of arrangements taking A in the first place.It will be 5!.Then CA,then CAI,CAL and so on

Soham Chanda - 8 years, 5 months ago
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