SATHEE: Permutations And Combinations Question 133

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

Show Answer

Answer:

Correct Answer: C

Solution:

The arrangement can be make 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) place. Hence required number of ways $ {{=}^{7}}C_4=35 $ .

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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:

Answer:

Solution:

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