moneydude242

Combinatorics is not my strongest asset, but I'll give it a shot...

I think we could do this question using complementary counting. In other words, we could find all of the combinations of sets that do not contain an ace, and subtract that from the total number of sets. If the piles are indistinguishable, then one pile would have the two of spades and the two of hearts, and you would do \({ 4 \choose 2 }\), to find the number of ways to find the other sets, which evens out to 6. Now, we need to find the total number of sets, which I think would be \(\frac{{6 \choose2} \cdot {4 \choose2} \cdot 1 }{6} = 15\), since you would first choose two numbers out of the six to be in the first set, then 2 numbers out of the four remaining for the second set, and since the third set is already decided, you would just multiply by \({2 \choose 2} = 1\). However, I divided by 6, since if the sets were indistinguishable, you would need to divide by \(3!\) in order to get rid of the repeats. (I am pretty skeptical about this step, if anyone could verify / help me with this, that would be greatly appreciated!) 

Finally, since there are 15 total cases, and there are 6 that do not work, there would be 9 cases that do work, so the final answer would be \(\boxed{\frac{3}{5}}\)

However, take whatever I am saying with a grain of salt, since as I said before, I am not the best at combinatorics... 

For the all the units digits under 9, it just won't work. Even if the first digit is divisible by 7, adding one to the units digits will only change it by 1 value, which doesn't make it divisable by seven. If you know modular arithmetic, the first number would be 0 mod 7, and the second 1 mod 7. For the value 9, it gets slightly more tricky. However, if you try a number like 759, adding 1 will give you 760, or -1 mod 7. Adding another 9, like 5599 (it there must be an even amount of 5 and 9s - can you see why?) pulled up to 5600 will give you 11. 555999 gives you 556000, or 16 (doesn't work!). The next one (55559999) is much too big. From now, we can consider numbers that end with 0. For the number 770 (769), it doesn't work. Same for for 7770, 77770, 777770, etc. 700 doesn't work, and 7000 doesn't. But 70000 works! 70000 and 69999 works! Sadly, afterwards no other number works. So your answer is one. (BTW, I'm not very sure if this is correct, because I somewhat brute forced this.)

Also, just a bonus thing I learned, the sum of the digits (if it has a nine - otherwise it doesn't work) has to be divsible by 21 (7*3) - idk who asked though

I would like to draw a diagram for you, but I was too lazy to learn Asymptote 

But I drew a lil' diagram on my paper, and found that 3 adjecent side are the \(10\) sides. Honestly, I don't have a proof for this, but I'm sure that this means the "rectangular" prism is actually a cube. That means that the side lengths are \(\sqrt{10}\), or the remainder is \(10\sqrt{10}\). 

This person isn't very good at combinatorics, but I guess I will give this a try? 

Let's call the American skaters \(a\), the Canadian skaters \(c\), and the other skaters \(o\). 

(Quick note, here are the numbers for \(a,c,\) and \(o \).)

\(a=2\)

\(c=5\)

\(o=12\)

Than, in the top 3 skaters, we want a standing that looks something like this: \((a_1,o_6,c_2)\)

(for this problem, I'm assuming that the Americans or Canadians are different, let me know otherwise...)

We can think just think of one case, since we will be multiplying the probablities, and multiplication is commutative.

(BTW, let me know if there is AT LEAST ONE in the problem, or just one American/Canadian)

I'm just gonna take the case \((a,c,o)\). For the American, there are 2 choices: \(a_1 \) and \(a_2\). For the Canadians, likewise, there are 5 choices. And for the other 12 people, there are 12 choices. We can simply multiply these 3 numbers together and get \(2*5*12 = 120\). We can multiply this number by 6, since there are \(3! \) ways to arrange the standings, so we get \(120*6 = \boxed{720}\)

First off, you want to simplify the equation.
\(3x^2 - 2x^2 = x^2 \)

\(12x+14x = 26x \)

\(4-18 = -14 \)

From this we can write out a more simplifed equation:

\(x^2+26x-14 = \text{__}(x+\text{__})^2 + \text{__}\)

From on here, we can use Simon's Favorite Factoring Trick (SFFT) to make the left side into a similar form as the right.

\((x+13)^2 = x^2 + 26x + 169, \text{ so if the third blank equals x, than } 169+x = -14, \text{or } x = -183\)

Then, you can factor the right hand side as

\((x+13)^2 -183\)

So that gives you your answer: the first blank is 1 (there is no coefficient, so that means it's one), the second is 13, and the third is -183.

OK corrected version:
 

In each of three boxes are four straws with integer lengths 1 cm, 2 cm, 3 cm, and 4 cm. A straw is randomly selected from each box. What is the probability that a triangle can be formed with the three straws chosen?

Could anyone also add an explanation with their answer? Thanks! 

ummm oops changing that now

Thank you very much for your explanation! I got the answer before this, but it is also nice to learn the equation!