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Apply the same principle to both rat and bird.  
 
Apply the same principle to both rat and bird.  
 
Hint: Look for the intersections, as long as they share same characters, there can't be any intersection between them. If they do not share any chars, they intersect, treat the 2 words as 2 blocks and remember the count the remaining characters. The total would just be 26! subtract the union of the 3 sets.
 
Hint: Look for the intersections, as long as they share same characters, there can't be any intersection between them. If they do not share any chars, they intersect, treat the 2 words as 2 blocks and remember the count the remaining characters. The total would just be 26! subtract the union of the 3 sets.
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Thanks for the help on 14! It made is so much clearer :)
  
 
Can someone please refresh my memory on when we do the over counting stuff... In the homework numbers 6 and 30 did not have any overcount right? Because the order really didn't matter? And then in 30 since there was replacement it didn't matter either?? Is that the right reasoning or no...
 
Can someone please refresh my memory on when we do the over counting stuff... In the homework numbers 6 and 30 did not have any overcount right? Because the order really didn't matter? And then in 30 since there was replacement it didn't matter either?? Is that the right reasoning or no...
  
 
[[Category:MA375Spring2010Walther]]
 
[[Category:MA375Spring2010Walther]]

Revision as of 17:20, 27 January 2010

Can anyone tell me how to start #61? Maybe I'm just missin something, but I can't figure it out.

Same here, I don't even understand what it(#61) wants.

Hint on 61: The problem wants you to count all the diagonals between vertices. There are none for triangles, 2 for squares and so on. For the n case we want to think of a way to count them. Here is the hint, remember that every vertex can't connect to itself or the two adjacent vertices. Use this to come up with a way to count how many diagonals each vertex can make, then adjust for over count.

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Can anyone tell me how to go about doing problem 14?

Can someone help me with #20d. Does it mean a positive integer under 1000 that is divisible by only the numbers 7 and 11? ^***YES***

It means the set containing numbers divisible by 7 union the set containing the numbers divisible by 11


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Can anybody help me out with 30h. Does the book mean only starts OR ends with BO or does it include the one the starts with AND ends with BO?

It wants to know starts OR ends with BO. Think of the equation for union for this problem. That being, # that start with BO + # that end with BO - # that start and end with BO.

I need some help with 44. Anyone know how to do it?

For 30h, "or" always means union

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For 44, 1)there are 6 different consecutive positions. 2)on those consecutive positions there could be 1 or 0. 3)there are 2^5 ways to fill other positions. 4)then combine the above ideas together, and it is over to you

Thanks for the help on 44. It really helped me out and now i understand how to get it.

In regards to the steps above for 44: depending on how you implement those ideas, be very careful of over count.

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anyone have any ideas on how to go about 7.5 #14???

For 14, try thinking of both letters and words as distinct objects that can be ordered. ___________________________________________________________________________________________________________________

Can anyone help with 20e from 5.1. Number 40c is like it. Whats it asking for when it wants EXACTLY ONE of something?

20e is asking how many positive integers less than 1000 are divisible by 7 or divisible by 11 but not divisible by both 7 and 11. Examples: 14 is divisible by exactly one of 7 and 11. 77 is not divisible by exactly one of 7 and 11. _______________________________________________________________________________________________________________

7.5 #14 For this question, I assumed that the alphabets cannot be repeated and there are 26! ways of arranging it anyway you like it. Then you should consider using the inclusion exclusion principle to look for the strings with fish rat or bird. So obviously there are 3 sets and you want to find the union of all these 3 sets. Those strings contain fish is one set, rat one set and bird another set.

Example: Take "fish" as one block, you are left with 22 characters and therefore there's 23! ways. Apply the same principle to both rat and bird. Hint: Look for the intersections, as long as they share same characters, there can't be any intersection between them. If they do not share any chars, they intersect, treat the 2 words as 2 blocks and remember the count the remaining characters. The total would just be 26! subtract the union of the 3 sets.

Thanks for the help on 14! It made is so much clearer :)

Can someone please refresh my memory on when we do the over counting stuff... In the homework numbers 6 and 30 did not have any overcount right? Because the order really didn't matter? And then in 30 since there was replacement it didn't matter either?? Is that the right reasoning or no...

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood