kris.yarmak123
Junior Member
- Joined
- Jun 13, 2019
- Messages
- 64
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...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?
Thanks.[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
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?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.
Exactly.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?
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.For #9, shouldn't I have just written "52! = 8,0658E67"? Like 52×51×50×49×48.....
I think that whoever wrote these questions is not well versed in writing counting questions,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?
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?[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
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 itExactly.
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.
I apologise for not answering for a long timeWith 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?
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.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
Got it, thank you for your helpFine 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.