Difference between revisions of "14" - Rhea

Latest revision as of 16:13, 28 January 2009

Ok does anyone know how to go about this problem?

I believe all you have to do for this problem is take the total number of ways to permute the 26 letters of the alphabet , which is 26!, and subtract all of the strings which contain the words 'fish', 'cat', and 'bird'. Namely if A='fish', B='cat', and C='bird' - |A or B or C| = |A| + |B| + |C| - |A and B| - |A and C| - |B and C| + |A and B and C|.--Spfeifer 19:06, 27 January 2009 (UTC)

However, if I understand the question correctly, each letter is only getting used once, so if 'i' is being used to make 'fish', it can't also be used to make bird. Same with 'r' for rat and bird. I'm i correct in thinking this? If so, using the notation from above, |A and C| and |B and C| should both be 0, correct?

Sounds like that is correct.. but I have no idea how to count the number of occurrences that has both fish and rat inside 26 length words.. I am bit comforted about the fact that I don't need to find lA and Cl nor lB and Cl.. but not quiet.. anyone has any idea?? --Kangw 22:37, 27 January 2009 (UTC)

--Here's a quick hint: try thinking of the strings "fish" and "rat" as single "letters." So for "fish" you will have: fish,a,b,c,d,e,g,j,k,l,m,n,o,p,q,r,t,u,v,w,x,y,z...or a total of 23 "letters" to arrange. --mkburges

-- According to above, Namely if A='fish', B='rat', and C='bird' - |A or B or C| = |A| + |B| + |C| - |A and B| - |A and C| - |B and C| + |A and B and C|. This formula is 23! + 24!(rat)+23!- 21!(fish and rat) -0(cause we can't have fish and bird)-0-0. Is it correct? -Eugene

-Yes, that would be the total for counting all the permutations that had fish, bird or rat in them. However, be sure to look again at the problem, because it asks how many do NOT have fish, rat, or bird in them. So, take the universe total (number possible without restrictions) and then subtract the number you just found. --mkburges