Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - 13.2 Permutations and Combinations - 13.2 Asses Your Understanding - Page 855: 26

Answer

The ordered arrangements: $123,124,125,126,134,132,135,136,142,143,145,146,142,153,154,156,162,163,164,165$ $213,214,215,216,231,234,235,236,241,243,245,246,251,253,254,256,261,263,264,265$ $312,314,315,316,321,324,325,326,341,342,345,346,351,352,354,356,361,362,364,365$ $412,413,415,416,421,423,425,426,431,432,435,436,451,452,453,456,461,462,463,465$ $512,513,514,516,521,523,524,526,531,532,534,536,541,542,543,546,561,562,563,564$ $612,613,614,615,621,623,624,625,631,632,634,635,641,642,643,645,651,652,653,654$ $P(6,3)=120$.

Work Step by Step

The ordered arrangements: $123,124,125,126,134,132,135,136,142,143,145,146,142,153,154,156,162,163,164,165$ $213,214,215,216,231,234,235,236,241,243,245,246,251,253,254,256,261,263,264,265$ $312,314,315,316,321,324,325,326,341,342,345,346,351,352,354,356,361,362,364,365$ $412,413,415,416,421,423,425,426,431,432,435,436,451,452,453,456,461,462,463,465$ $512,513,514,516,521,523,524,526,531,532,534,536,541,542,543,546,561,562,563,564$ $612,613,614,615,621,623,624,625,631,632,634,635,641,642,643,645,651,652,653,654$ We know that $P(n,r)=n(n-1)(n-2)...(n-k+1).$ Also $P(n,0)=1$ by convention. Hence $P(6,3)=6\cdot5\cdot4=120$.
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