Eight children are standing in a queue.
i) in how many ways can the queue be formed?
ii)how many arrangements are possible if the tallest child stands at the end of the queue?

Dear Student ,
Please find below the solution to the asked query :

1 Number of ways in which queue can be formed=8!=8×7×6×5×4×3×2×1=40320 ways .2 Number of arrangements , if tallest child stands at the end of queue=7!=7×6×5×4×3×2×1=5040 ways .
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