Arithmetic Ability
Permutations and Combinations

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Q.

The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ?

View Answer

Correct choice: C

Explanation:

Let 'Y' be the youngest player. The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways. The required number of ways = 3(3!) = 4320.
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