Coefficient of $t^{24}$ in $(1 + t^{2})^{12} (1 + t^{12}) (1 + t^{24})$ is

^{12} C_{6} + 3

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^{12} C_{6} + 1

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^{12} C_{6}

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^{12} C_{6} + 2

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Solution

The correct option is D $^{12} C_{6} + 2$
Here, coefficient of $t^{24}$ in ${{(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})}$
$=$ coefficient of $t^{24}$ in ${{(1 + t^{2})}^{12} . (1 + t^{12} + t^{24} + t^{36})}$
$=$ coefficient of $t^{24}$ in ${{(1 + t^{2})}^{12} + t^{12} {(1 + t^{2})}^{12} + t^{24} {(1 + t^{2})}^{12}}$
$($ neglecting $t^{36} {(1 + t^{2})}^{12})$
$=$ coefficient of $t^{24} = (^{12} C_{12} +^{12} C_{6} +^{12} C_{0}) = 2 +^{12} C_{6}$

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