Difference between revisions of "7.5 Homework MA375Fall2008walther" - Rhea

Revision as of 08:44, 11 September 2008

#12 in 7.5

It would be amazing if someone could explain the intersection between A and B on problem number 12. I have so far the squares being 31 and the cubes being 10. But I am not sure how to get the intersection. Any ideas? Also I do not understand 20 or 28 at all. If someone could please help me out that would ROCK! Thanks

Possible Solutions

I think that if you want numbers between 1 and 1,000 that are both cubes and squares you'd do (1,000)^1/6 but thats just my guess because when you want,

how many squares: (1,000)^1/2 = 31

how many cubics: (1,000)^1/3= 10

so how many squares and cubics: (1,000)^1/6 = 3

12. A: # of squares: 31 B: # of cubics: 10 A intersection B: 3

31+10-3=38

I also struggled with finding the intersection of the two, and although I got the same answer as you, I'm still lost. I need to know WHY you are choosing to raise 1000 to the power of $\frac{1}{6}$ .

Solution

We know that 1, 4, 9... are all squares, but we don't know where it stops. In other words, what number when squared, is the last square under 1000 (and thus we should stop counting)?

The answer to that question is $$ a^2=1000 $$ . It's clear now that $a=1000^\frac{1}{2}$ =31.(some decimal) and so the largest square under 1000 is $$ 31^2=961 $$ .

Similarly, the last cube under or equal to 1000 is 1000 itself, $$ 10^3 $$ .

Also:

$1000^\frac{1}{5}$ = 3

&

$1000^\frac{1}{6}$ = 3

How do you know to take 1000 to the power of $\frac{1}{6}$ and not $\frac{1}{5}?$

Solution

By inclusion-exclusion, we need to take out the numbers we counted twice (ie numbers which are squares and cubes). Because 3 is not divisible by 2, you can get a numbers which is a square and cube by squaring some integer, and then cubing that result. For example, 2 squared is 4, and 4 cubed is 64. This can be written symbolically as $$ (a^2)^3 $$ , where a is the integer being squared.

If you are like "ok, but how do I simplify that... struggling to remember exponent laws", just use the definition of exponent. No extra memorization required. $$ a^2=a*a $$ . and $$ b^3=b*b*b $$ . So $$ (a^2)^3 = (a^2)*(a^2)*(a^2)=(a*a)*(a*a)*(a*a) $$ . Since integer multiplication is "associative", you can forget about parentheses. So it just becomes $$ a*a*a*a*a*a $$ . There are 6 a there, so it equals $$ a^6 $$ .

Using a similar approach to above, $1000^\frac{1}{6}$ =3.something. (luckily, $1000^\frac{1}{5}$ also equals 3.something), so there are only 3 numbers counted twice.

#20 in 7.5

If anyone is still having trouble with 20, it might help to look at the answer to 19 in the back of the book. It gives an explicit formula for the number of elements in the union of 5 sets. It should help make the Principle of Inclusion-Exclusion a little more clear.

#28 in 7.5

Can someone rephrase the question or shed some light on what this question is asking? I looked at the solution for #29 which seems to be quite similar, but it was a notation we haven't learned in class.

I looked at the solution for #27, which seems to be a similar problem to #28. I think the inclusion/exclusion property for sets also holds for probabilities, and I think the point of the "no two events can occur at the same time" stipulation is to tell you that the probability of any intersection of events, whether 2, 3, or n, will be zero. That said, I think you just need to add the probabilities of the independent events.

^This is what I did, but I felt like it might have been too simple. Maybe when we get our homework back he'll go over a few problems?

I would like to see one of these done in class. If I could see a full example of one of these, I think it wouldn't be that hard.

I trouble with this as well. Maybe we could ask for more explanation in class on Tuesday

#14 in 7.5

How should we really start this question?

Am I missing something, like designated word length?

I think it means the whole 26 letters in succession like abcdeFISHgklmnopqrstuvwxyz

but i don't know where to go from there

-- I think so too... here is what I am thinking (i'm unsure): There are 26! ways to create a string of the 26 letters of the alphabet. There are only 22! ways for "fish" and "bird" to appear: FISHabc..., aFISHbc..., abFISHc... all the way to ...xyzFISH (along with all the rearrangements... bFISHac etc.)

Similarly with "rat", there are 23! ways for that to appear in the 26 letter string.

So, I think that the solution may be: 26! - 22!*2 - 23!. I may be totally off in my thinking though... Anyone else have ideas?

--I think that it is any arrangement of the 26 letters (abc, bca, acb) because the order of the letters would matter to form the words. The above idea I think is on the right track. There are 22! ways for the other letters to be ordered if "fish" or "bird" appears in the string. However, there are 23 different places that "fish" can be in that string. The other thing to consider is that "fish" and "rat" may both be in the same string on some occassions. Yet, "fish" and "bird" or "bird" and "rat" cannot both occur because one letter would be used twice (i in the first case and r in the second case). My thought for the solution is: 26!-23*22!-23*22!-24*23!+210. Any other ideas?

I had almost the same thing. However, I got that there are 420 ways for fish and rat to be together instead of just 210.

Quick Note* 23*22! = 23! and 24*23! = 24!

For the number of ways that fish and rat can exist in the same combination, I arrived at 21! ways. My line of thinking is that there are 26 letters, but with fish and rat you are removing 7 letters, leaving 19 letters. Now if you consider fish as a unit, rat as a unit, and the 19 remaining letters 1 unit each, there are 21! unique ways to arrange them. So I had my final answer as: (all combos) - (combos with fish) - (combos with bird) - (combos with rat) + (combos that we removed twice that contain both fish and rat): 26! - 23! - 23! - 24! + 21!