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Permutation under Restriction

The number of...

Question

The number of ways in which six + signs and four – signs can be arranged in a row so that no two – sings occur together is

A

6!7!

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B

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C

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D

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Solution

The correct option is D
The number of ways six + signs in a row = $\frac{6!}{6!} = 1$ way Number of gaps between six + signs = 7 Four – signs can be placed in 7 gaps = $\frac{^{7} P_{4}}{4!} =^{7} C_{4}$

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