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!
• B. $6{!}^7{C}_4$
• C. ${}^7{P}_4$
• D. ${}^7{C}_4$

Answer 1

The correct option is D ${}^7{C}_4$

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 =$\frac{{}^7{P}_4}{4!}{=}^7{C}_4$