I have a test coming up late next week or early in the week after next. In any case I want to be ready and so I have been practicing problems from the text a lot and just wanted to make sure I am not making any mistakes. If you see an answer that is wrong please let me know so I can try to see what I am doing wrong. For this next problem there is a club consisting of six distinct men and seven distinct women. 1) In how many ways can we select a committee of four persons so that Mabel and Ralph do not serve together? Since there are 13 people in the club there are C(12,4) possibilities for committees without Ralph and C(12,4) committees without Mabel. We want the union of the two sets. But we will end up counting the intersection twice and so we must subtract that off of our total. The intersection is where neither of the two are in the committee. That number is C(11,4). So we have 2*C(12,4)-C(11,4). Does that sound correct? 2) How many eight-bit strings contain exactly three 0's? An eight-bit string with 3 0's has 5 1's. We can start by placing the first 0. There are 8 possibilities for that, then there are 7 possibilities for the next and 6 for the last one. So there are P(8,3) possible 8 bit strings with 3 zeros. 3) How many eight-bit strings contain three 0's in a row and five 1's? Well since the 0's are all in a row we can place the first one and the other two will follow it. There are 6 places to start the first 0. So 6 is the answer 4) How many eight-bit strings contain at least two 0's in a row? There is 1 eight bit string with 8 1's and no two zeros in a row. There are 8 eight-bit strings with 7 1's and 1 0 and no two zeros in a row. Now for an eight-bit string with 6 1's there are exactly 7 places to stick the 2 0's so that no two zeros are together. In other word the 0's must go between the 1's. So for 6 1's there are C(7,2) and then using the same technique there are C(6,3) strings with 5 1's and then there are C(5,4) strings with 4 1's. So we add up all these bad cases and subtract them from the total cases to get the number of good cases. 2^8-[1+8+C(7,2)+C(6,3)+C(5,4)]=201 strings that have at least 2 0's in a row. For the next few questions I need to find the number of five card poker hands, selected from an ordinary 52-card deck, that have the properties indicated. 5) Containing four of a kind, that is, four cards of the same denomination. Well there are 4 suits of any domination. So we just need to pick the domination which can be done in 4 ways. Then for the remaining card there are 52-4=48 possible choices. So by the multiplication principle we have 4*48. 6) Containing all spades. There are 13 cards that are spades and so there is C(13,5) possible ways to have all spades. 7) Containing cards of exactly two suits. Well first off we choose the two suits which can be done in 6 ways. Once this is done there are 26 cards left to pick from So there are C(26,5) ways to pick cards from two suits. But we must not count the ways that have all of either suit we take away C(13,5) two times, once for each possible suit. So in total 6*[C(26,5)-2*C(13,5)]. 8) Consecutive and of the same suit.(assume the ace is the lowest denomination) 4 possible suits and 9 possible first cards so 4*9. 9) Consecutive (ace is lowest denomination). Well there are 9 choices for the first card and there are 4 choices for the suit of that card. Then the next 4 cards we pick the suit for each from 4 possible suits. So in total we have 9*4^5. 10) Containing two of one denomination, two of another denomination, and one of a third denomination. Well there are C(13,3) choices for the denomination. Then for two dominations that have two cards there are 6 choices for the two cards because there are 4 suits and we select 2. Then for the third denomination there are 4 choices. So in total there are C(13,3)*2*C(4,2)*4. I think that is enough to today, maybe tomorrow I will post the rest of the questions. Thanks for the help everyone.