A) 12
B) 10
C) 8
D) 6
Correct Answer: C
Solution :
\[{{T}_{4}}={{T}_{3+1}}\]\[=\,{}^{n}{{C}_{3}}\,\,\,{{a}^{n-3}}{{b}^{3}}\] Þ \[{}^{n}{{C}_{3}}=56\] Þ \[\frac{n!}{3\,!\,\,\,\left( n-3 \right)\,!}=56\] Þ \[n(n-1)(n-2)=56.6\] Þ \[n(n-1)(n-2)=8\,.\,7\,.\,6\] Þ \[n=8\].
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