How Many Different Combinations?

In summary: I don't really care about this crap. I just wanted to see if I could argue something to get a few points back.In summary, there are 5 questions, you must answer 3, and there are 10 possible arrangements of the 3 questions.
  • #1
sjaguar13
49
0
You are taking a test with 10 questions, you only have to answer 7, how many ways are there to do it?

10 x 9 x 8 x 7 x 6 x 5 x 4 ?

How many different ways are there to answer the questions if you must answer at least 3 from the first 5 questions?

5 x 4 x 3 x 7 x 6 x 5 x 4 ?
 
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  • #2
sjaguar13 said:
You are taking a test with 10 questions, you only have to answer 7, how many ways are there to do it?
10 x 9 x 8 x 7 x 6 x 5 x 4 ?
Not.

Look up the difference between permutations and combinations.
Which one applies to this question?
at least 3 from the first 5 questions?
What are the possible scenarious for "at least 3"?
For each one of them, how many different ways are there to pick remaining questions?
 
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  • #3
Isn't a permutation ordered? It shouldn't matter if you do question 5 before question 1.

What about 5 choose 3 x 7 choose 4?
 
  • #4
Reducing numbers often helps.

Try solving a "smaller" question first, listing all possible arrangements first, and comparing their number to one you calculated.
--------------
You are taking a test with 5 questions A, B, C, D, E, you only have to answer 3, how many ways are there to do it?
---------------
How many different ways are there to answer the questions if you must answer at least 2 from the first 3 questions?
 
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  • #5
Alright, I have no idea. If we go really small and just go for 2 from the first 3, that is:
AB
AC
BC

3. 3 questions, choose 2 is 3.

3! is 6. Divide by 2! (2) is 3. That doesn't seem right.

I don't even know what 3 choose 2 is, but I am guessing that isn't right either. I looked up permutations on Google and it used factorials.

n!/(n-k)! would be 3!/1! which is 6.

Can you give me a bigger clue?
 
  • #6
Let's deal with the easier one first.
ivybond said:
You are taking a test with 5 questions A, B, C, D, E, you only have to answer 3, how many ways are there to do it?
Is the order in which you answer questions important?
In other words, is answering questions A, B, C the same as C, B, A?
I hope I understand the question right and rephrase is correctly:
How many ways are there to choose a set of three letters out of a set of five {A, B, C, D, E}?
Is it 5P3 or 5C3?
Why?

By hand:
ABC
ABD
ABE
ACD
...
CDE
.
How many altogether?

Sorry, the second question is more involved, and I have to go, but you should try to write out all possible arrangements, just like above.
My reservation is that it's phrased somewhat vaguely:
sjaguar13 said:
How many different ways are there to answer the questions if you must answer at least 3 from the first 5 questions?
How many questions do you need to answer? Can a number of answered questiones be any number from 3 to 10 inclusively?
Sjaguar, did you copy this question exactly or posted it from your notes/memory?
 
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  • #7
It was a test question. It's posted exactly. How I answered them was what I typed before the ? Those are wrong, though, because I got a big red X through it and -10 on that question. There's no explanation as to why it was wrong.

Is it supposed to be 10C7?
 
  • #8
sjaguar13 said:
It was a test question. It's posted exactly. How I answered them was what I typed before the ? Those are wrong, though, because I got a big red X through it and -10 on that question. There's no explanation as to why it was wrong.

Is it supposed to be 10C7?

Yes, that's right.

Start working with the smaller numbers and make a list of subsets, number of which you are asked to find.
(I showed a partial list for one example).

It's too bad your teacher did not write any comments or suggestions.
Maybe you could find some book with a concise chapter on the basics of combinatorics. I believe Barron's SAT II Math 2 has one.
 
  • #9
I don't really care about this crap. I just wanted to see if I could argue something to get a few points back.
 
  • #10
But don't you want to understand why the answer is that ? Because there's every likelihood that you're going to get it wrong again and have to argue with the teacher again.

You cannot solve the second part without a slightly deeper understanding of what's going on. Would you like that, or don't you care enough ?
 
  • #11
sjaguar13 said:
I don't really care about this crap. I just wanted to see if I could argue something to get a few points back.

FWIW: You wouldn't have to nitpick points if you tried harder to understand the principles involved and you would save yourself the embarrassment of having to grovel. Furthermore, you could have pride in knowing that you actually earned a better grade.

Just my tuppence.
 
  • #12
I'm taking 6 classes and working. There are some things I just don't care about, this happens to be one of them. As for the second part, I got:
5C3 x 5C4 + 5C4 x 5C3 + 5C5 x 5C2

The teacher actually didn't grade the tests. A teaching assistant did. Partial credit was supposed to be given, but it wasn't done really well. My teacher said if you feel like you should have gotten more points, you could write down the questions and submit them. I had to do it by today. It may be good to understand the principles, but I have deadlines. I can't wait around trying to derive formulas.

I would have pride even if I bought a better grade. This class is stupid, and I hate it.
 

What is the concept of "How Many Different Combinations?"

The concept of "How Many Different Combinations?" refers to the number of possible arrangements or groupings that can be made from a given set of elements.

How do you calculate the number of different combinations?

The number of different combinations can be calculated using the formula nCr = n! / r!(n-r)!, where n is the total number of elements and r is the number of elements in each combination.

What is the difference between combinations and permutations?

Combinations are a way to calculate the number of different groupings without considering the order, while permutations take into account the order of the elements in the groupings.

Why is understanding different combinations important in science?

In science, understanding different combinations is important for designing experiments, analyzing data, and predicting outcomes in complex systems.

Can the number of different combinations ever be infinite?

No, the number of different combinations is always finite, as it is dependent on the number of elements in the set and the number of elements in each combination.

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