The fundamental counting principle and permutations math 30-1

[kris.yarmak123]

Hello. We have just started a new unit on permutations and counting principle, and I am stuck ok these two questions? Can someone please explain me how do I solve #7,9?

 

[Dr.Peterson]

Please explain the thinking that led to what you wrote for these problems, so we can discuss with you where that thinking may be wrong.

For #7, you wrote two things, each of which is wrong in a different way!

For #9, why did you start with 1?

 

[kris.yarmak123]

[MATH][/MATH]

Please explain the thinking that led to what you wrote for these problems, so we can discuss with you where that thinking may be wrong.

For #7, you wrote two things, each of which is wrong in a different way!

For #9, why did you start with 1?

well, for 7 I wanted to write 3×2×1=6 but I was not sure if it's correct. So I wrote some weird solutions which are wrong...
For 9,I already understood that I have to start with the biggest possible amount of choices. So it should have probably been 5×4×3×2×1=120

 

[Dr.Peterson]

[MATH][/MATH]well, for 7 I wanted to write 3×2×1=6 but I was not sure if it's correct. So I wrote some weird solutions which are wrong...
For 9,I already understood that I have to start with the biggest possible amount of choices. So it should have probably been 5×4×3×2×1=120

Thanks.

In #7, we have to arrange 3 letters from the 6 letters of FRIDAY. Your 3×2×1=6 suggests that there are 3 ways to choose the first letter, 2 ways to choose the second, and 1 for the third. Is that right? How many choices are there for the first?

In #9, we are dealing 5 cards from a deck of 52. your answer of 5×4×3×2×1=120 suggests that there are 5 possible cards to put in the first place. Is that true? Aren't there 52 cards to choose from?

After you have that settled, you'll have to realize that a "hand" doesn't care what order it is held in. All the other problems are permutation problems (arranging); this is a combination problem (merely choosing). Let us know what you have been taught about combinations. (It's possible that you haven't been taught that at all yet, and this question doesn't belong here unless you think of a hand in the sense of the order in which it is dealt.)

 

[kris.yarmak123]

Thanks.

In #7, we have to arrange 3 letters from the 6 letters of FRIDAY. Your 3×2×1=6 suggests that there are 3 ways to choose the first letter, 2 ways to choose the second, and 1 for the third. Is that right? How many choices are there for the first?

In #9, we are dealing 5 cards from a deck of 52. your answer of 5×4×3×2×1=120 suggests that there are 5 possible cards to put in the first place. Is that true? Aren't there 52 cards to choose from?

After you have that settled, you'll have to realize that a "hand" doesn't care what order it is held in. All the other problems are permutation problems (arranging); this is a combination problem (merely choosing). Let us know what you have been taught about combinations.

For #7, i think for first letter there are 6 choices, but for the second it's 5 choices , and for the third there are 4 choices, so shouldn't I have stopped at 6×5×4=120?
For #9, shouldn't I have just written "52! = 8,0658E67"? Like 52×51×50×49×48.....

 

[Dr.Peterson]

For #7, i think for first letter there are 6 choices, but for the second it's 5 choices , and for the third there are 4 choices, so shouldn't I have stopped at 6×5×4=120?

Exactly.

For #9, shouldn't I have just written "52! = 8,0658E67"? Like 52×51×50×49×48.....

That would be true if you were arranging all 52 cards. But you only need 5. Therefore, if we suppose that order matters, the answer will be 52×51×50×49×48 = 311875200.

Now please let us know whether you have heard of combinations at all. That is the right way to solve #9 as I'd define "hand". It amounts to continuing from where we are, and observing that there are 5! ways to arrange each 5-card hand, so we have to divide by 5! to get the final answer.

 

[pka]

Hello. We have just started a new unit on permutations and counting principle, and I am stuck ok these two questions? Can someone please explain me how do I solve #7,9?

I think that whoever wrote these questions is not well versed in writing counting questions,
#9 "How many five card hands can be dealt from a fifty-two card deck?"**
As written that means ten hands of five each. I am quite sure that is not what is meant.
I suspect that it means "A hand of five cards is dealt from a standard deck. How many such hands are possible?"
This question is about content as opposed to 'order'. As such it is a combination rather that permutations.
Then \(\displaystyle \dbinom{52}{5}=~?\)

As ** is written the answer is \(\displaystyle \prod\limits_{k = 0}^9 {{\dbinom{52-5\cdot k}{5}} } \) SEE HERE

 

[Steven G]

[MATH][/MATH]
well, for 7 I wanted to write 3×2×1=6 but I was not sure if it's correct. So I wrote some weird solutions which are wrong...
For 9,I already understood that I have to start with the biggest possible amount of choices. So it should have probably been 5×4×3×2×1=120

With your method, 5×4×3×2×1=120, you are saying that whether you have 7 cards or 52 cards or 12,000 cards there are EXACTLY 120 different 5 cards hands. Does that sound correct to you? Do you think that the number of cards you start with should matter?

 

[kris.yarmak123]

Exactly.

That would be true if you were arranging all 52 cards. But you only need 5. Therefore, if we suppose that order matters, the answer will be 52×51×50×49×48 = 311875200.

Now please let us know whether you have heard of combinations at all. That is the right way to solve #9 as I'd define "hand". It amounts to continuing from where we are, and observing that there are 5! ways to arrange each 5-card hand, so we have to divide by 5! to get the final answer.

On math, we have only done some applications like "how many arrangements can 9 people make?" and he told us how to solve it - 9×8×7×6×5×4×3×2×1= x or 9!=x that was only intro so I'm only getting into it

 

[kris.yarmak123]

With your method, 5×4×3×2×1=120, you are saying that whether you have 7 cards or 52 cards or 12,000 cards there are EXACTLY 120 different 5 cards hands. Does that sound correct to you? Do you think that the number of cards you start with should matter?

I apologise for not answering for a long time

 

[Steven G]

On math, we have only done some applications like "how many arrangements can 9 people make?" and he told us how to solve it - 9×8×7×6×5×4×3×2×1= x or 9!=x that was only intro so I'm only getting into it

Fine you are new at this but you need to think reasonably. It is one thing not to know how to do a problem or to make certain mistakes but the mistake you made meant that you were not thinking about the consequences of what you tried. I try many things that are not correct but I catch many of them because they are not reasonable.

 

[kris.yarmak123]

Fine you are new at this but you need to think reasonably. It is one thing not to know how to do a problem or to make certain mistakes but the mistake you made meant that you were not thinking about the consequences of what you tried. I try many things that are not correct but I catch many of them because they are not reasonable.

Got it, thank you for your help