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Prove that the number of ways in which n books can be placed on a shelf when two particular books are never together is (n – 2) × (n – 1)!. |
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Answer» Regarding the two particular books as one book, there are (n – 1) books now which can be arranged in n – 1Pn –1, i.e., (n – 1)! ways. Now, these two books can be arranged amongst themselves in 2 ! ways. Hence the total number of permutations in which these two books are placed together is 2 ! . (n – 1)!. The number of permutations of n books without any restriction is n !. Therefore, the number of arrangements in which these two books never occur together = n !–2 ! .(n – 1) ! = n . (n – 1) ! – 2 . (n – 1) ! = (n – 2) . (n – 1) ! |
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