If $P = 21 (21^{2} - 1^{2}) (21^{2} - 2^{2}) (21^{2} - 3^{2}) . . . (21^{2} - 10^{2})$ , then $P$ is divisible by

22!

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21!

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19!

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20!

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Solution

The correct options are
B $21!$
C $19!$
D $20!$
$P = 21 (21 + 1) (21 - 1) (21 + 2) (21 - 2) . . . (21 + 10) (21 - 10)$
$= (21 - 20) (21 - 9) . . . (21 - 1) 21 (21 + 1) (21 + 10) . . . (21 + 10)$
$= 41 \times 40 = 11$
which is divisible by $21!$ , and hence by $20!$ and $19!$ .

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