Drishti answer this query

Dear Student,
Solution) 3 particular numbers can be put in r places in (r-2) ways and 3 numbers can be arranged themselves into 3! ways. Now, (n-3) numbers will be arranged in (r - 3)  places in ${}^{n-3}P_{r-3} ways.$
Thus, using the fundamental principle of counting, the req. number of permutations will be 3! . (r-2) . ${}^{n-3}P_{r-3} ways.$
Regards!
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