Easy -1.2 This is a straightforward two-part question requiring direct application of the binomial coefficient formula and binomial theorem. Part (a) is pure recall/calculation of 6C3 = 20, and part (b) requires identifying that the coefficient is 6C3 × (-2)³ = -160. Both are routine textbook exercises with no problem-solving or insight required, making this easier than average.
0 for just 20 seen in second part; M1 for \(6!/(3!3!)\) or better
\(-160\) or ft for \(-8 \times\) their 20
2
Condone \(-160x^3\); M1 for \([-]2^3\times[\text{their}]\ 20\) seen or for \([\text{their}]\ 20\times(-2x)^3\); allow B1 for 160
## Question 17:
$20$ | **2** | 0 for just 20 seen in second part; M1 for $6!/(3!3!)$ or better
$-160$ or ft for $-8 \times$ their 20 | **2** | Condone $-160x^3$; M1 for $[-]2^3\times[\text{their}]\ 20$ seen or for $[\text{their}]\ 20\times(-2x)^3$; allow B1 for 160 | Total: **4**
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