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Question

5th terms from the end in the expansion of (x32−2x2)12 is

A
7920x4
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B
7920x4
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C
7920x4
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D
7920x4
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Solution

The correct option is C 7920x4
Tr+1=nCr(a)nr(b)r
Now,
(x322x2)12
5th term, from the end has 8 before it =T8+1
T8+1=12C8(x32)4(2x2)8

T8+1=12×11×10×91×2×3×4×2824x4

T8+1=12×11×10×924×16x4

T8+1=11×5×9×16x4

T8+1=7920x4=7920x4



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