Coefficient of a32 in the expansion of(a4−1a3)15 is
A
15C4
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B
−15C4
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C
0
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D
15C3
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Solution
The correct option is B15C4 Tr+1=nCran−rbr Applying to the above question we get Tr+1=(−1)r15Cra60−4ra−3r =(−1)r15Cra60−7r ...(i) For the coefficient of a32 60−7r=32 28=7r r=4 Substituting in (i) we get T5=15C4a32 Thus the required coefficient is 15C4