Counting

[AuSmith]

Are these good problems? Any suggestions? Should I just quit? Ok well, here are some more. The difficulty level should increase.

(1) Mr. Prather, a hungry man, treats himself to some seafood. He can order either lobster, crab, or seaweed. The restaurant has 4 lobsters, 6 crabs, and 6 seaweeds. If Mr. Prather predetermined that he will eat 9 lives, how many choices does he have?

(2) The beings on Filo's planet Zarcon are not quite as chromosomally oriented as us, resulting in a people of little variety. There are only 4 genders. Any two Zarconites of the same gender are truly identical. It's as if they all jumped off four assembly lines. Ok, well if they don't have chromosomes, how do they reproduce? Luckily, the planet is already made of organic material. So, whenever one of the Zarconites feels biologically compelled to reproduce, it flips its 4-sided coin, which tells the Zarconite which of the 4 genders to make. The Zarconite obediently follows the corresponding reproduction code so impressed in its being, scooping up flesh much in the same way that a small child builds a sandcastle. All that to say, there are only 4 kinds of Zarconites. Here's the problem:

My pet skunk ran off the porch and died, so I headed off to the nearest Zarconite sale. With such a low demand, the salesman only brought 2 of each gender (total of 8). In how many ways can I purchase a total of 4 Zarconites?

(3) Can you relabel the six sides of two ordinary dice and keep the distribution of rolls? (2=1/36 of the time, 3=1/18 of the time, etc.)(You can't just rearrange the faces. Find something other than {1,2,3,4,5,6}.)

[fht]

3) {0,1,2,3,4,5} and {2,3,4,5,6,7}
1) if you're not looking for 16P9, then it's probably 4+5+6+7+6=28.
They are interesting problems, pray keep them coming.
Best Wishes

[AuSmith]

1) Right, I was thinking how many ways could Mr. Prather have ordered. When you order lobster, you don't say, I want Hoppy and Bob. Rather, you ask for some nonnegative integer number of lobsters. 28 is correct. Sam split the problem into 5 cases; 5 for the number of lobsters that were ordered. Then, for each case, find how many ways there are to order crab. Once lobster and crab are picked, the number of seaweeds is determined already so that the total number of lives reaches 9.

lobsters--- crabs
-------------------------------------------------
0---------- 3,4,5,6
1---------- 2,3,4,5,6
2---------- 1,2,3,4,5,6
3---------- 0,1,2,3,4,5,6
4---------- 0,1,2,3,4,5

Hence, 4+5+6+7+6


3) Okay, okay. How about keeping it to positive integers? I meant to say that.

[fht]

3) {1.3.4.5.6.8}with{1.2.2.3.3.4}

The sums are 2.3.3.4.4.5/4.5.5.6.6.7/5.6.6.7.7.8/6.7.7.8.8.9/7.8.8.9.9.10/9.10.10.11.11.12
Count the distributions, you'll get 2*1.3*2.4*3.5*4.6*5.7*6.8*5.9*4.10*3.11*2.12*1

I haven't tried to prove whether this is the only solution, maybe it is.
Best Wishes

[AuSmith]

Good. How'd you find the labeling?

[fht]

I went for a lil walk and thot about it meanwhile. I knew thr had to be some balance or symmetry in the chgs. And I considered several diff't kinda chgs, and just.... thot of the solutn.

I'm not tryg to be rude or anythg, but it's the truth. It was really nothg systematic, lol.

[AuSmith]

[AuSmith]

[suchenzang]

2) I'm not sure if I did this right, but here goes nothing:

There are three cases that we need to consider:
AABB - where you have two pairs of identical Zarconites = 4C2 = 6
ABCC - where you have one pair and two other Zarconites = 4C2 x 2 = 12
ABCD - where they are all different = 1

So... 6+12+1=19 different ways

That is, if you can't differentiate between skunks of the same gender...

[AuSmith]

Yea, that's correct.

All of these problems I meant to accommodate generating functions. For this one, you can find the coefficient of [unparseable or potentially dangerous latex formula] in

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[unparseable or potentially dangerous latex formula]

Both ways aren't too icky.