A) 15
B) 18
C) 35
D) 42
Correct Answer: C
Solution :
The arrangement can be make as \[2{{b}_{n}}=\frac{n}{^{n}{{C}_{0}}}+\frac{n}{^{n}{{C}_{1}}}+......+\frac{n}{^{n}{{C}_{n}}}\], the \[(-)\] signs can be put in 7 vacant (pointed) place. Hence required number of ways\[{{=}^{7}}{{C}_{4}}=35\].
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