Hence the multiplication axiom applies, and we have the answer (4P3) (5P2). For every permutation of three math books placed in the first three slots, there are 5P2 permutations of history books that can be placed in the last two slots. So the answer can be written as (4P3) (5P2) = 480.Ĭlearly, this makes sense. Therefore, the number of permutations are \(4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 = 480\).Īlternately, we can see that \(4 \cdot 3 \cdot 2\) is really same as 4P3, and \(5 \cdot 4\) is 5P2. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. Therefore, to calculate the number of combinations of 3 people (or letters) from a set of six, you need to divide 6 by 3. Now, there are 6 (3 factorial) permutations of ABC. In Combinations ABC is the same as ACB because you are combining the same letters (or people). The fourth slot requires a history book, and has five choices. So ABC would be one permutation and ACB would be another, for example. Since the math books go in the first three slots, there are 4 choices for the first slot,ģ choices for the second and 2 choices for the third. We first do the problem using the multiplication axiom. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books? You have 4 math books and 5 history books to put on a shelf that has 5 slots. Since two people can be tied together 2! ways, there are 3! 2! = 12 different arrangements The multiplication axiom tells us that three people can be seated in 3! ways. Let us now do the problem using the multiplication axiom.Īfter we tie two of the people together and treat them as one person, we can say we have only three people. So altogether there are 12 different permutations.
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