1. In how many different ways can the numbers ‘256974’ be arranged, using each digit only once in each arrangement, such that the digits 6 and 5 are at the extreme ends in each arrangement ?
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By: anil on 05 May 2019 01.39 am
Case 1 : 6 at left end and 5 is at right end : 6 _ _ _ _ 5 Now, four empty places can be filled by 2,9,7 and 4 in = $$4!$$ ways = $$4 imes 3 imes 2 imes 1 = 24$$ Case 2 : 6 at right end and 5 at left end : 5 _ _ _ _ 6 Similarly, no. of ways = $$4!$$ = $$4 imes 3 imes 2 imes 1 = 24$$ $$ herefore$$ Total no. of ways = $$24 + 24 = 48$$
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