I was doing this problem and now I am second guessing myself. Could you let me know if I'm right? I just want to make sure. Thank you in advance.
There are 8 women and 9 men trying out for a cheerleading squad.
The squad will consist of 6 people, only 2 men.
How many squads are possible?
you find the possible combinations of 2 men from the 9 trying out:
you find the possible combinations of 4 women from the 8 trying out:
you multiply those possibilities together (by fundamental counting principle):possible squads.
Am I missing anything?
For problems like these, I always liked this method. Its similar to bob bundy's method, and will show you more steps.
3x² + 7x + 4.
1. Multiply the outside numbers 3 * 4 = 12
2. Find two factors of 12 that add up to 7.
In this case, the two factors of 12 that add up to 7 are: 3 and 4 or;
3 + 4 = 7 and
3 * 4 = 12.
Rewrite your original equation replacing the middle term
3x + 4x, since 3x + 4x = 7x.
So instead of...
3x² + 7x + 4, we now have...
3x² + 3x + 4x + 4.
factor out the 3x from first two terms, and the 4 from the second two...
3x(x + 1) + 4(x + 1).
Now pull out the (x + 1) and you have....
(x + 1)(3x + 4)
Here is an idea for you. I have not figured out how to generalize it, but hopefully it will put you on the path you are looking for.
As you increase the number of sides of a regular polygon, it looks more and more like a circle.
So, you can choose a regular polygon, put it on an xy-plane, then rotate it around the y or x axis. You can use the inscribed circle of the regular polygon as your guide for the spherical shape of your cubes.
If you make the side lengths 1 of your regular polygon, then it is not too far of a stretch to picture cubes when rotating around the y-axis.
Here is an example using a regular ocatgon(see attached image)
For any regular polygon, with a side length of 1, the radius of the inscribed circle for n sides is...
for an octagon...
and the circumference of the inscribed circle of an octagon will be...
The arc length between the center points of two of your cubes on this circle is.
So you know that for approximately each .948 of the circumference of a circle, you will have one cube.
With a little work, you can find
and the circumference of the circle after rotation, with D1 as its diameter will be
this tells us that you will be able to fit 5 cubes around this part, with .622736878 of circumference left that you can divide evenly between all of the cubes so spacing is equal.
The middle diameter is the same as the circle and will fit 8 cubes around it.
So, when its all said and done, after rotating around the axis of rotation, you will have 1 cube on the top, 5 under it, 8 in the middle, 5 under that, and 1 on the bottom, giving you a total of 20 cubes using this method.
I hope I was clear and gave you some ideas as to how to approach this!
Hi gAr and bobbym;
I approached this problem by looking at the ways the race could finish, based on the number of ties possible.
For 1 horse, 0 ties are possible.
For 2 horses, 0 ties and 2 ties are possible.
For 3 horses, 0, 2, and 3 ties are possible.
For 4 horses, 0, 2, 3, and 4 ties are possible. etc.
The number of ways that the places(1st, 2nd, 3rd, 4th, etc.) can be assigned to all the horses, ties included, has given me this triangle.
1 1 1
1 2 1 1
1 2 2 1 1
1 3 3 2 1 1
1 3 4 3 2 1 1
1 4 5 5 3 2 1 1
1 4 7 6 5 3 2 1 1
1 5 8 8 7 5 3 2 1 1
1 5 10 11 10 7 5 3 2 1 1
1 6 12 15 13 11 7 5 3 2 1 1
1 6 14 18 18 14 11 7 5 3 2 1 1
Lets look at the 4th row, when 4 horses are racing.
The first number tells us that there is 1 way to assign the finishing places of the race if there is 0 ties (1 2 3 4), which has 4! = 24 possible outcomes.
The second number tells us that there are 2 ways to assign the finishing places of the race if 2 horses tie (2 2) and (2 1 1) which has
2!/2!(4!/2!2!) + 3!/2!1!(4!/2!1!1!) = 1(6) + 3(12) = 42 possible outcomes.
The third number tells us that there is 1 way to assign the finishing places to the race if 3 horses tie (3 1) which has 2!/1!1!(4!/3!1!) = 2(4) = 8 possible outcomes.
The fourth number tells us that there is 1 way to assign the finishing places to the race if 4 horses tie (4) which has 1!/1!0! = 1 outcome.
Which gives us a total of 24 + 42 + 8 + 1 = 75 outcomes.
I was wondering if one of you two can notice a pattern in the triangle that will give us the next line.
I did notice bobbym that you did some research and noted that a closed form is not likely.
Would this indicate that there is not a pattern that we could find to predict the next line in this triangle?
Hi Learn Everything;
Here are some of my thoughts on this....
I think the three essentials you need in life are food, water, and shelter. Maybe a few more depending on who you are...
Everything else is not needed "to exist" or "absolutely essential""
Social studies, english, math, science, everything else, all the things you learn in school, you do not need to be alive.
Is it really "absolutely essential" to know a language? You could live in the middle of nowhere and sustain yourself without ever speaking a word. People will always be able to communicate, whether or not they understand each other, just so they can survive.
It all comes down to what decisions you want to make that will make you the person who you want to be...
Life is all about time. You are born and you die; it begins then it ends(as we know it). There is a set amount of time that you live your life. The more you know, the more you can do with your life. You make better decisions and get a better quality of life if you are a knowledgeable, quality person.
If you want to learn math and know how to solve things on a piece of paper before you spend all the time, money, and resources to do it in real life, then that is who you are...
If you want to be the person that spends 10-30 seconds measuring something with a floppy tape measure... that is who you are.
If someone who knows the Pythagorean Theorem can figure out the length of their roof by going into their attic, measuring the length of the floor and the height to the pitch, that is who they are.
If the other guy wants to just measure it the "old fashioned" way, bust out the scaffolding, and risk his life climbing onto the roof... that is who he is. (Though I wouldn't recommend this method right now where im at since theres about 5 inches of snow on the roof!)
What type of decision maker do you want to be? Do you want to be the guy in the attic or the one on the roof?
It seems like you are making a decision in your life right now that will define "who you are." If it is what you want, and you are to learn math in order to accomplish it, so be it!
I never knew that knowing what a coach purse was would have a positive impact on my life. But, low and behold, I saw a cutie and struck up a conversation over her coach purse. She was impressed I knew the brand.
You also never know when the math you learn will show up in your life. But, when it does, you will appreciate that part of your life a lot more... Once again... its just the type of person who you want to be.
All in all....
Having logical and mathematical knowledge is much bigger than a few real life examples, it is a way of being. It can be a part of you or not... it is your decision.
What's it in the end? Its the quality of life!
aha... very interesting! Expansion of the mind again!
Could an induction proof flow like this somewhat? (I haven't done higher level math in a while and I am trying to remember what I did in college.)
We are trying to prove that:
s^2 = c^3 or
s = (c^3)^(1/2).
The solution sets (c, s) are of the form (n^2, n^3) for n = 1, 2,....
Indeed, when n = 1,
we have the solution set (1, 1) and 1 = 1^2 = 1^3 = 1.
In general, substituting c = n^2 and s = n^3 into s = (c^3)^(1/2)
gives us n^3 = ((n^2)^3)^(1/2) = n^3.
Does n + 1 work?
(n + 1)^3 = (((n + 1)^2)^3)^(1/2) = (n + 1)^3
Indeed it does, so there are an infinite amount of s and c that satisfy the condition s^2 = c^3.
use the quadratic formula for u^2 - 2yu - 1 = 0. Make sure you use a = 1, b = -2y and c = -1 when you substitute.
You will see that y + sqrt(y^2 + 1) comes out as one of the solutions for u, hence
u = y + sqrt(y^2 + 1)
when you substitute e^x back in for u, you then have to take the ln of each side which will:
convert e^x to x on the left and create ln(y + sqrt(y^2 + 1)) on the right.
Then simply switch the x and the y to find the inverse since that has not been done yet, and you have the problem in the form that you want now.
84 * 76
would be (a + b) (a - b) = a^2 - b^2
or (80 + 4) (80 - 4) = 80^2 - 4^2 = 6400 - 16 = 6384.
This does require a more "specialized" starting position with the numbers that are given to me, but I shall remember this one as well!
I have tried (a + b + c)^2, but I think I would need more of a ram upgrade in my brain in order to do these in my head. Nonetheless, I always found it interesting how these expansions can make our lives easier!
i just did something with my class they thought was interesting,
the monty hall problem...
It touches on theoretical/experimental probability, outcomes, events, and can provide a very valuable lesson on assuming things based on what you see.
I started off telling a story about Marilyn Vos Savant and how the monty hall problem came to be in her life. The fact is that she answered the question correctly yet so many mathematicians, doctors, and professors wrote her and told her to correct her mistake because they thought she was wrong!
I tell the kids that they are going to solve the problem today and be able to understand something that all of those doctors, mathematicians, and professors could not!
The problem is like this:
The teacher is the host, a student comes up to play the game.
There are 3 doors in front of the classroom. Behind 2 of the doors are goats, and behind one of the doors is a car. The student picks one of the doors. (I used playing cards, where (2) 2's were the goats and an A was the car)
The host then opens one of the doors that has a goat behind it since he knows what is behind all of the doors.(they picked one card, i looked at the other two and showed a 2)
Then the host gives the student a choice: stick with the same door you orginally picked, or switch to the other one that hasn't been opened.
The question here is: Should you stick, or should you switch, or does it not matter?
Before reading any farther, if you have not heard this problem before, think of how you would answer this question. Everyone who I have talked with this about has thought the same thing initially.... See if you do too!
The same thing that everyone thinks is: it doesn't matter if you stick with your original choice or switch to the other card since there are only two cards left and it is a 50 50 chance you win. Most people will stick with their card when given the choice to change(this might be an interesting psychological question!)
The fact is,
this is the wrong choice, since when you change you choice you have a 2/3 chance of winning the car!
At this point, I go around the room and have everyone play the game and we tally the results based on our classroom playing the game( I try to have 20-30 trials).
A great way to do this is online with a monty hall simulator:
but you can do it the long way too. Just keep track of wins and losses for keeping your choice and changing your choice. (good for block scheduling)
On the above website you can also run 1000's of simulations with keeping choice and changing chioce as well.
This would be considered the experimental probability of this situation. You can find it to be 67% chance of winning when you change your choice by running a couple thousand simulations, same for keeping your choice comes out at 33%. The law of large numbers can also be explained here, since in the beginning of the experiment when the number of trials was low, the percentages of wins and losses were not what they theoretically should be.
The mistake most people make when assuming that it is just a 50 50 chance at the end is they forget that there was 3 original choices you could have made. If you keep your choice, then you will only have a 1/3 chance of winning, because you are sticking with your original 1 of the 3 cases all the way. When you change it becomes different.
Think about it, each time the host has to show a goat. You have a 2 out of 3 chance of picking one of those goats in the beginning. So if you pick one of those goats initially, and the host shows you the other one, then when you change you will win the car. The only time you lose when changing is when you pick the car initially, which is a 1 out of 3 chance. So you win 2 out of 3 when changing and you win 1 out of 3 when keeping your original choice.
I have them fill out a diagram like this to do the theoretical probability of this situation.
The three original choices you could make are:
Car Goat 1 Goat 2
Stick Change Stick Change Stick Change
Win Car Win Goat Win Goat Win Car Win Goat Win Car
There are 6 possible outcomes, 3 you stick, 3 you change. Of the 3 when you stick you win a car 1/3 times. Of the 3 when you chage, you win a car 2/3 times.
All in all, it is the better choice to change since you have a 2/3 chance of winning when you do so. You can talk for a while about these types of things all while having them engaged.
Its called the monty hall problem because of its similarity to the lets make a deal game show.
Here is something interesting that I have been using to amaze my 8th graders~!
I have them pick a two digit number and I will square it in my head...
Here is how I do it...
(a + b)² = a² + 2ab + b²
Say they pick 24; 24 is (20 + 4); a = 20 and b = 4 so...
so 20² + 2(20)(4) + 4² = 400 + 160 + 16 = 560 + 16 = 576.
This works for all two digit numbers.
Look at 49 (this ones a little tougher to do in the head but if you practice, you will master!)
40² + 2(40)(9) + 9² = 1600 + 720 + 81 = 2320 + 81 = 2401
Your post on whether or not 26 was the only number jammed between a square and a cube intrigued me.
My investigation has only brought me more questions.... here goes.
Well, I approached this problem like this:
can I find an x and a y, such that:
(a) x^2 + 1 = y^3 - 1 OR (b) x^2 - 1 = y^3 + 1.
I solved (a) for y to get y = (x^2 + 2)^(1/3) and similaraly (b) became y = (x^2 - 2)^(1/3).
I set the domain for natural numbers in excel and plugged these equations in.
(a), from x = 1 to x = 10,000 returns only 1 rational solution (5, 3) and indeed
5^2 + 1 = 26 = 3^3 - 1
(b), from x = 1 to x = 10,000 returned only 1 rational solution (1, -1) and indeed
1^2 - 1 = 0 = (-1)^3 + 1.
So if we look at both sides of the coin, there appears to be two numbers "jammed" between a square and a cube: 26 and 0.
Now this does suggest that if you want the square and cube to be natural numbers, then 26 is the only solution.
But if you say the square and the cube can be integers, then 26 and 0 are solutions.
Furthurmore, if we say the square and cube can be irrational numbers, then we have infinite solutions as the functions show us. An example would be the solution (sqrt(6), 2) for (a) where (sqrt(6))^2 + 1 = 7 = 2^3 - 1.
Another interesting tidbit: as x got very high, the difference between the (a) and (b) functions seemed to be approaching zero. Maybe the + 1 and - 1 become negligible as the values of x^2 and y^3 get into the tens of millions and they will not impact the value of x^2 and y^3 enough to cause for the situation we are looking for.
This is far from a proof, but my investigation i feel has offered some great insights as to how we may formally prove this situation. It seems though that this is the theorem we would like to prove:
Prove that, for all natural numbers, 26 is the only number between a square and a cube, or
Prove that, for all integers, 26 and 0 are the only numbers between a square and a cube.
Let me know what you think!