Question

Find the total number of ways in which six '+' and four '–' signs can be arranged in a line such that no two '–' signs occur together.

Answer 1

Total number of '+' signs = 6

Total number of '–' signs = 4

six '+' signs can be arranged in a row in $\frac{6!}{6!}$ = 1 way [$\because$ All '+' signs are identical]

Now, we are left with seven places in which four different things can be arranged in $7_{p_4}$ ways but all the four '–' signs are identical, therefore, four '–'signs can be arranged in $\frac{7_{p_4}}{4!}=\frac{7!}{\frac{(7-4)!}{4!}}=\frac{7!}{3!\times 4!}$

= $\frac{7\times 6\times 5\times 4!}{3\times 2\times 4!}=7\times 5=35$

Hence the required number of ways

= 1 $\times$ 35 = 35