Random math questions

[TrueTears]

Is the intersection of finitely many closed sets closed? I'm aware it is for a 'collections of sets' but I'm not sure if this means that it has to be an infinite collection or it can be finite.

Yes. In fact, it holds even stronger than that, the intersection of any arbitrary collection (could be infinite) of closed sets is closed. The proof is very straightforward, just a simple application of De Morgan's laws and the relation between open/closed sets.

[#1procrastinator]

How would you evaluate this without using trigonometric or hyperbolic substitutions? Also for (I tried u=x^2+a^2 and ended up with an integral of the second form)

[kinslayer]

How would you evaluate this without using trigonometric or hyperbolic substitutions? Also for (I tried u=x^2+a^2 and ended up with an integral of the second form)

Then

[Zlatan]

A florist has to make a floral arrangement. She has 6 Banksias, 5 wattles and 4 Waratahs. All the flowers of each kind are different.In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind ?

Thanks

[kinslayer]

A florist has to make a floral arrangement. She has 6 Banksias, 5 wattles and 4 Waratahs. All the flowers of each kind are different.In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind ?

Thanks

Combinatorics isn't my strong point, but let's go anyway.

By the multiplication principle, there are: ways to choose the first nine spots of the floral arrangement.

There are 6 flowers left to choose from, and there is a different floral arrangement for each flower chosen. We've already counted the number of ways these flowers can be included in the initial 9, so the total number of arrangements is simply

e: fixed mistake

[Zlatan]

Ummm .... I'm not too familiar with Combinatorics either. So could you explain it in better detail ? Thanks anyways for this answer anyway !!!!!

[kinslayer]

Ummm .... I'm not too familiar with Combinatorics either. So could you explain it in better detail ? Thanks anyways for this answer anyway !!!!!

No problem, I am basically just using combinations:

http://en.wikipedia.org/wiki/Combination

and the multiplication principle:

http://en.wikipedia.org/wiki/Rule_of_product

I looked at the first nine spots because it's simply the product of combinations.

There are ways to choose 3 banksias out of 6 banksias, then there are ways to choose 3 wattles out of 5 wattles, then there are ways to choose 3 waratahs out of 4 waratahs.

Since each way of rearranging the individuals results in an entirely new arrangement, you must multiply all of them together to get the total number of arrangements with 9 spots. But now you have one more spot with six remaining flowers. We don't need to distinguish between the type now because there aren't any further restrictions and each one will result in a new arrangement. So we just multiply the previous result by 6 to get the final answer.

[#1procrastinator]

To solve the constant coefficient PDE , the method my lecturer used was to make the following substitutions so that all but the ith partial derivative disappear




How do you work out these substitutions?