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 ?
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
= $$4 \times 3 \times 2 \times 1 = 24$$
Case 2 : 6 at right end and 5 at left end : 5 _ _ _ _ 6
Similarly, no. of ways = $$4!$$
= $$4 \times 3 \times 2 \times 1 = 24$$
$$\therefore$$ Total no. of ways = $$24 + 24 = 48$$
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