Combinations

This is a question from an assignment given in a Computational Algorithm class.

Ten research labs are attempting to determine the efficacy of proposed drugs in treating cancer. All the drugs are ordered from a pharmaceutical company (all ordered from the same one). Each of the labs want to order 1 drug to test. The choices are the following: Xenr, Phum, and Tsela.

The questions:

a.) How many orders are possible?

b.) Xenr appears promising. How many orders are there where at least 3 labs order Xenr?

c.) The pharmaceutical company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fill?

I am not sure which combinatoric principals would be used. a.) apppears it'd just be 10^3.

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Originally Posted by fifthrapiers

This is a question from an assignment given in a Computational Algorithm class.

Ten research labs are attempting to determine the efficacy of proposed drugs in treating cancer. All the drugs are ordered from a pharmaceutical company (all ordered from the same one). Each of the labs want to order 1 drug to test. The choices are the following: Xenr, Phum, and Tsela.

The questions:

a.) How many orders are possible?

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as you said, the answer is 3^10

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b.) Xenr appears promising. How many orders are there where at least 3 labs order Xenr?

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three labs order Xenr, there are seven more labs that can order 3^7 different combinations. Therefore there are 3^7 combinations

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c.) The pharmaceutical company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fill?

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Find the number of possible combinations where 3 or more labs by Tsela, which is 3^7 (which equals 2187)

Now subtract that from the total possible combinations: 59049-2187=56862

so there are 56862 possible combinations if no more than 2 labs can buy Tsela.

make note: 3^10=59049

Thanks for the reply; I thought my answer (and subsequently yours), appeared too simple! Apparently this is not the correct answer, according to the professor. She said she could understand how that answer was created, and said to add,

"All of the 10 labs are @ the same univ. and the univ. places the "drugs" request @ the pharmac. company."

According to the professor, I should be using the r-selection method for this; that is, order is not important, and repetition is allowed. I believe it'll use the equation C(r+n-1, r)..

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Originally Posted by fifthrapiers

Ten research labs are attempting to determine the efficacy of proposed drugs in treating cancer. All the drugs are ordered from a pharmaceutical company (all ordered from the same one). Each of the labs want to order 1 drug to test. The choices are the following: Xenr, Phum, and Tsela.
The questions: a.) How many orders are possible?
b.) Xenr appears promising. How many orders are there where at least 3 labs order Xenr?
c.) The pharmaceutical company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fil.

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I can tell you that this is a very poorly worded problem!
I say that as someone who has been of many editorial boards, regional and national, for contest questions.

In this case, I would have suggested this revision: “The university’s buying agent has to order drugs from one pharmaceutical company. The agent is buying one of enr, Phum, and Tsela for each of ten labs on campus.
a.) How many orders are possible?
b.) Xenr appears promising. How many orders are there where at least 3 labs order Xenr?
c.) The pharmaceutical company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fill?
Now your instructor’s suggestion makes perfect sense!

Indeed, well not to me any way :D.

Tried working on c.): is it (10 choose 9) * (9 choose 3) ?

c.) The Pharmaceutical Company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fill?

If we take that to mean “How many orders from the 10 labs can the company fill with no more than 2 Tselas in the order” then that is the complement of at least three Tselas in the order. [C(N,k) means N choosing k.] The number of total possible orders is C(10+3-1,10)=C(12,10). The number of total possible orders with at least three Tselas is C(7+3-1,7), think of going ahead putting three in the ‘Tsela box’ and distributing the remaining seven in all three boxes. Thus, C(12,10)-C(9,7) is the total number of orders with no more than 2 Tselas.

Okay, I thought I had a and b right thinking c was too, but now it just confused me with your answer. Initially, I thought a.) would just be 3^10. I mean, there are 10 ways to choose the 1st drug, then 10 ways for the 2nd, and then 10 ways for the 3rd.

And for b, with 7 remaining labs Quick's answer of 3^7 seemed logical, but there has to be some combinatorics, so something's wrong.

It is fairly clear from what your instructor said that she is expects you to use the multi-selection rule: C(N+k-1,k) ways to make N choices from k different items. You are ordering ten from three different drugs. The labs are not a consideration here. We are simply interested in how many different orders are possible of ten from three.