HSC Stuff > HSC Mathematics Extension 1
3U Maths Question Thread
Happy Physics Land:
--- Quote from: cajama on February 06, 2016, 07:26:52 pm ---Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??
a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)
--- End quote ---
Hey Cajama:
Firstly I just wanna to assure you here that they are not typos, Im fairly certain that the questions are correct despite the answer does seem quite ridiculous. Good job on answering part a), remembering to divide by 2! which shows that you have recognised that there is a repetition.
Now for part B, I would like to invite you to draw just four underline dashes, each representing a number that can be placed in that spot. (i.e. __ __ __ __)
On the first underscore dash, there are only 4 numbers that can be placed there， which are 4, 4, 5, 6 (you cannot select 3 as the first digit because the number has to be greater than 4000). On the second underscore, there are also only 4 numbers that can be placed there (because a number, we dont know which one, has already been selected, leaving us with 4 options to choose from.) On the third underscore, we will only have 3 options to choose from because we have already selected two of them from all the numbers that we are provided with. And finally on the last underscore we will only have 2 options to choose from because we have already selected 3 numbers from all the numbers that the question provides us with.
Hence, the total number of different arrangements will seem to be 4 x 4 x 3 x 2 = 96 ways. BUT, WE HAVE NOT YET ACCOUNTED FOR THE REPETITION OF 4 IN THE NUMBERS THAT WE ARE PROVIDED WITH. Hence similar to what you have done in part a), it is necessary for you to divide 96 by 2! and you will obtain an answer of 48.
Ok, now we are up to part c). This one will be extremely similar to part b). I would recommend you to follow what I did in part b) and attempt part c) yourself to check whether you have truly understood what I have done. I will still post the solution below, but have an attempt on yourself.
So, similar to part b), draw four underscores __ __ __ __, each representing a digit.
On the first digit, we can only have three options of numbers, because from all the numbers we are given, only three of them will suit the criteria of "a number that is less than 5000". These three numbers, evidently, are 3, 4, 4. The second digit will have 4 options for us to choose from, because there is a total of 5 numbers which we are provided with and we have already selected one. The third digit is the same, we will have 3 options to choose from because we have already chosen 2 numbers from the 5 for the first two digits. And for the final digit, we will have 2 options to choose from because we have already chosen 3 numbers for the first 3 digits.
Hence the total amount of numbers that we can make below 5000 will be (3 x 4 x 3 x 2) / 2! = 36 (2! is to account for the repetition of 4 in the numbers that we are provided with)
Anyways great questions, hope you have understood my solution. Don't hesitate to post more questions if you need further assistance!!! :)
Best Regards
Happy Physics Land
Happy Physics Land:
--- Quote from: nerdgasm on February 06, 2016, 07:52:10 pm ---I think there is a typo; 40000 and 50000 should be what is written in the question (given the answers provided), or else every 5-digit arrangement would be larger than both 4000 and 5000.
Assuming that 40000 and 50000 are respectively meant:
For part b), I find it easier to look at which arrangements are not counted, because there are fewer cases to consider. In this case, only those arrangements starting with a 3 will be not counted. So, let's assume the starting digit of the arrangement is 3, so we have 3 _ _ _ _.
We then have to place 4, 4, 5 and 6 into the remaining four slots. This can be done in 4!/2! = 12 ways, by a similar argument to what you did in part a). All other arrangements work, so the total number of successful arrangements is 60 - 12 = 48.
For part c), we can again use the same 'counting the complement' technique. This time, we exclude those arrangements starting with 5 or 6. By the same working out as in part b), we can deduce that if the starting digit is 5, there are 12 arrangements (as you then have 5 _ _ _ _ with 3, 4, 4, 6 to place), and there are also 12 arrangements if the starting digit is 6 (6 _ _ _ _ , with 3, 4, 4, 5 to place). Hence, the total number of arrangements less than 50000 is 60 - 12 -12 = 36.
It is also possible to count the arrangements that are included for part c), but you do have to be a bit careful: if the starting digit is 3, then there are 12 successful arrangements. However, note that if the starting digit is 4, then you have 4 _ _ _ _ with 3, 4, 5, 6 to place. Because the remaining digits are all distinct, there are 4! ways to place them, and hence there are 24 successful arrangements. Therefore, the number of successful arrangements is 12 + 24 = 36, which matches the result we obtained earlier.
--- End quote ---
Hmm ok this is quite interesting. We have actually defined the question differently, you saw it as all the numbers that can be made and are greater than 4000 or less than 5000 and I saw the question as all the four-digit numbers than are great than 4000 or less than 5000. Im not too sure about the wording of this question or maybe is it just a typo of 40000 and 50000.
jakesilove:
--- Quote from: cajama on February 06, 2016, 07:26:52 pm ---Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??
a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)
--- End quote ---
Great answers all! Glad this forum is developing so quickly, and there are so many people willing to jump in and help others out!!!
Jake
nerdgasm:
--- Quote from: Happy Physics Land on February 06, 2016, 07:57:26 pm ---Hmm ok this is quite interesting. We have actually defined the question differently, you saw it as all the numbers that can be made and are greater than 4000 or less than 5000 and I saw the question as all the four-digit numbers than are great than 4000 or less than 5000. Im not too sure about the wording of this question or maybe is it just a typo of 40000 and 50000.
--- End quote ---
You're definitely right. the suggested answers work just as well if the question is considering 4-digit arrangements. I think that your interpretation is more likely to be correct and that there was no typo. Good job on your explanation too, I thought it was really clear!
*(The following is probably not part of the 3U maths course)*:
I think that the reason why we were able to get the same numerical answers despite our different interpretations is as follows:
Let us define a finite 'pool' of n members (the members are not necessarily distinct). I claim that the number of different n-member arrangements is the same as the number of (n-1)-member arrangements.
My idea is as follows: I propose a mapping from the set of n-member arrangements to the set of (n-1)-member arrangements. This mapping basically chops the last member off the n-member arrangement to get an (n-1)-member arrangement.
To show that this mapping is injective (or one-to-one), I use proof by contradiction: assume we have two distinct n-member arrangements that map to the same (n-1)-member arrangement. Then, it follows the two n-member arrangements must agree in their 1st, 2nd ... (n-2)th and (n-1)th positions. But because the pool is finite, it follows that there is only one member in the pool left to make the nth position. Therefore, the two n-member arrangements must in fact be the same.
To show that this mapping is surjective (or onto), note that for any (n-1)-member arrangement, there must be exactly one member from the original pool that was not used. Stick this member onto the end to form a n-member arrangement, which will map to the (n-1)-member arrangement.
Therefore, as this mapping is both injective and surjective, it follows that there are exactly the same number of n- and (n-1)-member arrangements from a finite pool of n members.
I think this explanation also explains our similarity in answers to parts b) and c) as well: in part b), since the starting digit (to be excluded) is 3 in both of our interpretations, we are essentially trying to place 4, 4, 5 and 6 into a 3-member arrangement or a 4-member arrangement in order to work out how many to exclude, which can be done in exactly the same number of ways. And similarly in part c), we are trying to place 4, 4, 5 and 6 into a 3- or 4-member arrangement if the starting digit is 3, and we are trying to place 3, 4, 5 and 6 into a 3- or 4-member arrangement if the starting digit is 4, both of which can be done in exactly the same number of ways.
foodmood16:
Hey I have a test coming up and I am having trouble with ext locus and parabola. The question is
Find the equation of the chord of contact AB of tangents drawn from an external point (x1, y1) to the parabola x2 = 12y
Thankyou :)
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