Engineering
Mathematics
Properties of Binomial Coefficients
Question
The coefficient of x7 in the expansion of (1 – x – x2 + x3)6 is :
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Solution
(1 – x – x2 + x3)6
(1 – x)6 (1 – x2)6
(6C0 – 6C1x1 + 6C2x2 – 6C3x3 + 6C4x4 – 6C5x5 + 6C6x6) (6C0 – 6C1x2 + 6C2x4 – 6C3x6 + 6C4x8 +............+ 6C6x12)
Now coefficient of x7 = 6C1 6C3 – 6C3 6C2 + 6C5 6C1
= 6 × 20 – 20 × 15 + 36
= 120 – 300 + 36
= 156 – 300
= – 144 Ans.
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