Permutations And Combinations Question 133
Question: Six ‘+’ and four ‘-’ signs are to placed in a straight line so that no two ‘-’ signs come together, then the total number of ways are [IIT 1988]
Options:
A) 15
B) 18
C) 35
D) 42
Correct Answer: C The arrangement can be made as $ 2b _{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) places. Hence required number of ways $ {{=}^{7}}C_4=35 $ .
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Answer:
Solution:
BETA
