Discussion

The number of sequences in which 7 players can throw a ball, so that the youngest player may not be the last is -.

Answer: C

Count all sequences and remove the forbidden ones.

\[ \text{total} = 7! = 5040 \]

\[ \text{youngest last} = 6! = 720 \]

\[ 5040 - 720 = 4320 \]

Equivalently, the last place has 6 acceptable players and the rest arrange freely: \( 6 \times 6! = 4320 \).

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