You could approach it by viewing it as three decisions and then multiplying the number of options for each decision together. There are 6 candidates for the first song, which leaves 5 possibilities for the second song and 4 possibilities for the third song. That creates 6 × 5 × 4 = 120 possible versions of the CD.
Assignment 6 © Princeton Review Management, L. L. C. | 191 Alternatively, you could use the permutation formula to get the same result. There are 6 candidates, so n = 6. Elmo is choosing 3 songs, so r = 3. When you plug these numbers into the formula, you get P n n r = − = = × × × × × × × = ! ( )! ! ! 6 3 6 5 4 3 2 1 3 2 1 120 . A third possibility is to break this problem into two sub-problems: a combination and a permutation. It’s not the most efficient way to answer the questions but it will provide some insight into the way combinations and permutations fit together. This understanding can be critical for answering more difficult questions that do not fit nicely into the patterns you’ve seen up to now. You can analyze Elmo’s decision in two stages: 1. He must decide which 3 songs he wants to include. This is a combination because order is not involved. 2. Once he has chosen the 3 songs, he must decide in what order to record them. This is a permutation because order is involved. In the first phase, you find the number of 3-song groups (combinations) that are possible. Then, you multiply by the number of ways to arrange a given 3-song group. That will give you the total number of permutations. # of combinations of 3 songs # of ways to order a group of 3 songs Total # of pemutations × = The first step is to determine how many 3-song combinations are possible. You could use the group formula for this: C n r n r = − = = × × × × × × × × × × = ! !( )! ! ! ! 6 3 3 6 5 4 3 2 1 3 2 1 3 2 1 20 . The 20 possible groups include ABC, ABD, ABE, and so forth. Next, suppose that Elmo has selected his 3 most favorite songs. He needs to determine the order in which to record these songs on the CD. This a permutation in which you choose 3 out of 3 items and arrange them. There are 3 × 2 × 1 = 6 choices. You could also use the permutation formula, with n = 3 and r = 3. You will get the same result. (Remember that 0! = 1.) So, there are 20 possible groups (combinations), each of which could be arranged in 6 different ways, for a total of 20 × 6 = 120 permutations. The key insight about combinations and permutations is that a permutation is just a combination that has been arranged in some order. Conversely, a combination is just a permutation with the ordering factored out. That’s why there is an extra r ! in the denominator of the combination formula. P n n r = − ! ( )! C n r n r P r = − = ! !( )! !
GMAT MANUAL 192 | © Princeton Review Management, L. L. C. C IRCULAR P ERMUTATIONS Questions involving circular permutations are very rare. In fact, it’s fine to skip this section. It’s only worth spending time on this topic if you’ve mastered everything else.
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