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Question

The coefficient of x7 in the expansion of (1xx2+x3)6. is?

A
132
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B
144
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C
-132
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D
-144
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Solution

The correct option is D -144
(1xx2+x3)6=(1x)6(1x2)6=(16C1x+6C2x26C3x3+6C4x46C5x5+6C6x6)(16C1x2+6C2x46C3x6+)
coefficient of x7 is
6C1.6C36C3.6C2+6C5.6C1=6×2020×15+6×6=144

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