I am having a tough time with these proof, I am not sure how to approach this one along with the other ones. Can someone help me at least start in the right direction with these?
For this I followed an example in the book and said (a) 7!, (b) 6! etc; but I wasn't sure how do to parts (e) or (f). For (e) since B is at the end of both of the required strings, does the required string turn into CABED? I guess my confusion is if the order with in the string matters. --Rhollowe 16:49, 4 February 2009 (UTC)
Rhollowe, you are correct! For e, because the strings share a common letter they must be combined into one string. so the answer would be 4!. For f, the answer is 0 because you cannot have two of the same letter in the permutation. And to who ever posted the initial question, the basic formula for doing these problems is: (total letter choices - letters used in the strings + number of strings)! So for part a it would be (8-2+1)!=7!=5040 Hope that helps! ---Kristen 19:52, 4 February 2009 (UTC)
Since there is only one page to this (does someone know how to fix this?) I'll post the solution to 22 (a) from section 5.4:
So in the LHS you are choosing r elements from a set of n elements AND further choosing k elements from that set of r elements. I thought of this as saying that you are essentially choosing just r from n. The LHS proves my RHS by saying that you choose k elements from a set of n elements AND a set of every element r from n that is NOT in the set of k. --Chris Ruderschmidt
