Question 18

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Jambmaths question: 

For what value of n is n + 1C3 = 4(nC3)?

Option A: 

4

Option B: 

6

Option C: 

3

Option D: 

5

Jamb Maths Solution: 

$\begin{align}  & ^{n+1}{{C}_{3}}=4\left[ ^{n}{{C}_{3}} \right] \\ & \frac{(n+1)!}{(n+1-3)!3!}=4\left[ \frac{n!}{(n-3)!3!} \right] \\ & \frac{(n+1)!}{(n-2)!3!}=\frac{4n!}{(n-3)!3!} \\ & \frac{(n+1)!}{(n-2)!}=\frac{4n!}{(n-3)!} \\ & \frac{(n+1)!}{n!}=\frac{4(n-2)!}{(n-3)!} \\ & \frac{(n+1)n!}{n!}=\frac{4(n-2)(n-3)!}{(n-3)!} \\ & n-1=4(n-2) \\ & n+1=4n-8 \\ & 3n=9;\text{   } \\ & n=3 \\\end{align}$ 

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