I have three questions that I'm super confused on. Answer one or all would be WONDERFUL. Thank you a lot. 1. In how many ways can 15 books to dividied by 3 people? 2. Camera club has 5 members, and math club has 8 members. Only one member of the public to both clubs. In many ways how can a committee of 4 persons is formed by at least 1 member from each group? 3. A soccer team played seven games and won for them, no bond. There are 35 settings of 4 wins and 3 losses. How much of this arrangement will the team have at least two wins in a row?
1. In how many many ways 15 books that can be shared by 3 people? There are several interpretations to this question: The most common are: a number of books for every person, and every book and every person different. (Eg Alice gets # 1 # 7, # 13, Bob got # 2 - # 6 and # 10 - # 12, Carol got vs. the rest. Switch Alice and Bob's books or take two and Carol give each for Alice and Bob.) Perhaps the most simple: How many ways to divide the 15 books in 3 sets of 5:? and there are many others in one of. For now I, will last only do simple 10C5 * 15C5 * 5C5 (= 15C5 * 10C5 * 1) is the number of ways to divide the book into 3 groups of 5 For the general case entirely, here is a outline: How to partition number 15 to 3 numbers are: from 15,0,0 to 5,5,5. 27 (or 19, if everyone gets at least one book) Then for each one of them, we have to choose who gets the book, by calculating the combination and then permuting them among 3 people. 2. Camera club has 5 members, and math club has 8 members. There is only one common to both club members. In how many ways can a committee of 4 persons is formed by at least 1 member of each group of clubs? Suppose Bob is on both. There are 11 other people. Two cases: Bob is on the committee, or not: a) Bob include: First vote for him, and then select any 3 of the 11 others, because both clubs are represented by Bob: 11C3 way to do it:. b) Bob exempted Now start with 11C4 (committee composed of others 11) =. Reduce 4C4 = 1 all the way with 4 others from the camera club, and 7C4 35 ways with 4 other people from the math club:. 11C3 + 11C4 - 1 - 7C4 = 165 + 330 - 1 - 35 = 459 Or we can say there 12C4 committees total (= 495 = 165 + 300) do not worry Bob, and then just subtract 1 which is 4 members of camera clubs who are not Bob and 35, the same for math club. 3. A soccer team played seven games and won for them, not there are ties. There are 35 settings of 4 wins and 3 losses. How much of this arrangement will the team have at least two wins in a row? The arrangement only that * not * have 2 wins in a row is: WLWLWLW So the answer is 34:. (Or the hard way WW times WLWW times 5C2 = 10 3C1 = 3 WLWLWW WLWLLWW 1 times 1 times WLLWLWW WLLLWWW WLLWW times 2 times 1 times 1 times SPL 4C2 = 6 LWLWW times 2 times 1 times LWLLWW LWLWLWW a LLWW * 3C2 = 3 LLWLWW LLLWW times 1 times 1 34 Total)
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