I have a few I'm having trouble with, but let's start with the easiest just so I can see if I'm even on the right track.

A set S has 12 elements. How many subsets of S have at most 4 elements?

T1

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- January 24th 2007, 12:26 PMTainted1Combinations question
I have a few I'm having trouble with, but let's start with the easiest just so I can see if I'm even on the right track.

A set S has 12 elements. How many subsets of S have at most 4 elements?

T1 - January 24th 2007, 12:43 PMThePerfectHacker
- January 24th 2007, 01:44 PMTainted1
Thank you for the response. I'm sorry, math isn't my strong suit. I'm just filling a requirement. I don't understand. We're working on combinations now like if I wanted just to take four out of the 12 it would be C(12,4). I just can't seem to figure out how to find how many have at most four.

Here's another one I'm killing myself over:

Select three coins from the following: 5 dimes, 3 nickels and 2 quarters.

The select 3 from 10 is simple C(10,3), but here's the real question.

In how many possible ways can the selection be made so that the value of the coins is at least 25 cents?

My last one is: Assume that the student has a cup with 11 writing implements: 6 pencils, 4 ball point pens, and 1 felt-tip pen.

1. In how many ways can the student select 7 writing implements?

Simple, C(11,7) = 330

2. In how many ways can the selection be made if no more than one ball point pen is selected?

I can get parts of the answer, but can't seem to put it all together.

T1 - January 24th 2007, 02:16 PMPlato
TPH gave you the answer to at most four: none+one+two+three+four.

In your notation C(12,0)+ C(12,1)+ C(12,2)+ C(12,3)+ C(12,4).

Just add all five numbers. That is the answer.

Select three coins from the following: 5 dimes, 3 nickels and 2 quarters.

The select 3 from 10 is simple C(10,3), but here's the real question.**This not correct unless we assume all the coins are distinct.**Are they? Usually we consider nickels are identical.

In how many possible ways can the selection be made so that the value of the coins is at least 25 cents?

List the possibilities: {n,n,n}, {n,n,d}**are the only set that do not have a value of at least 25 cents**. So we must count all the others.

You need to list all sets of three that work. - January 24th 2007, 03:04 PMTainted1
Thank you, both. I was thrown off by the notation. In the end I had made a simple arithmetic error at the beginning.

The coins solution was 104. I did exactly what you wrote, Plato, but got mixed up on the second step. The first C(10,3) all the coins were non-distinct. However, once into the second step it went N1N2D1, N1N2D2 and so on.

Now I just need to figure out the writing implements.

Thanks, I really appreciate it.

T1 - January 24th 2007, 03:14 PMPlato
Is there anything in your text material/book to indicate that we are to assume that the coins are all distinct?

If not, that is the nickels are identical, then here is the list of the way to choose three coins: [n,n,n], [n,n,d], [n,n,q], [n,d,d], [n,q,q], [n,d,q], [d,d,d], [d,d,q], [d,q,q]. That is just nine possible samples.