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In how many ways can 5 children be arranged in a line such that (i) two of them, Rajan and Tanvy, are always together? (ii) two of them, Rajan and Tanvy, are never together? |
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Answer» Correct Answer - (i) 48 (ii) 72 Total number of ways of arranging 5 children in a row = 5! = 120. (i) Considering Rajan + Tanvy as one child, this one and 3 more can be arranged in 4! = 24 ways. Rajan and Tanvy can be arranged among themselves in 2! = 2 ways. Required number of ways `=(24 xx2)=48`. (ii) When Rajan and Tanvy are never together, the number of arrangements `=(120-48)=72.` |
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