Recent content by SneakyArab

Alright, thank you so much!

Oh hey that makes sense. So maybe (14! * (15 choose 5) * 5!) ?

But then there could be a child in an even space, then two adults, then a child in an odd space. Wouldn't that mess things up?

Edited my first post. The problem doesn't state that they have to stay together, so assume they don't. Our professor who made the problem is out of town right now, and will be until after it is due, so I can't ask him for clarification.

Homework Statement Seven married couples, five of which with a single child, go to a movie theater. In how many ways can they line up to buy tickets so that no two children stand next to each other? This is the problem, word for word. The problem does not state that families have to be...

Ooooh ok, I see. so its 7 choose 2 * 5 choose 2 * 8^3 Thank you so much!

Homework Statement How many integers from 0-9,999,999 have exactly two 3's and two 5's as digits. Homework Equations I'm not really sure... The Attempt at a Solution The answer is 107520, if I'm not mistaken. I made a program to count it up for me, so I'm fairly sure that that is...

I think I have a working proof on this one now. Thanks for getting me started Slider!

seems I've found a classmate. I'm not sure if you mean the last symbol or next to last, so after the equation it reads: "is not rational" so its not a rational number.

Ah yes, back to basics for me. Ok so it seems that all possible values of f(x) are: -6,-5,-4,-3,-2,-1,0; thus range of f(x) = (-6,0). Correct?

Homework Statement I have to find the range of f(x) where: f(x)=7*floor(x) - floor(7*x) x is a real number. Then I have to prove that this set of numbers is correct. The Attempt at a Solution I am still on the range part, and thus am only asking about that part at the moment. It...

right, I had a couple of laws floating around in my head that I just didnt put. Thank you so much.

Ok I think I see now. So do I not need to do induction at all? Can I write the proof like this: For all n>2: there exists p prime: n<p<n! Proof: For all p<n: p | n! //p divides n! n! and n!-1 are relatively prime => n! and n!-1 share no common divisors => there must a prime p > n : p | n!-1

I don't know :( I feel stupid that I still don't know what that leads to.

Thanks. It doesn't have to be done by induction by the way, its just that that is one of the only proof methods that I have been taught thus far, and seemed like the most likely.