Question

Six$‘+‘$and four $‘-‘$signs are to be placed in a straight line so that no two $‘-‘$ signs come together, then the total number of ways are?

• A. $15$
• B. $18$
• C. $35$
• D. $42$

Answer 1

The correct option is C

$35$

Explanation for the correct answer:

Finding the required number of ways:

Number of $+$ signs = $6$

Number of $-$ signs = $4$

Given, the signs are arranged in a straight line so that no two $‘-‘$ signs come together.

This arrangement can be given as,

$-+-+-+-+-+-+-$

The above arrangement shows that there are $7$ places and the four $‘-‘$ sign can take any of the above places.

Hence, the favourable outcomes $=7$

Thus, the required number of ways $={}^{7}C_4$

$$\begin{gathered} =\frac{7!}{4!(7-4)!}\left[\therefore {}^{n}C_r=\frac{n!}{r!(n-r)!}\right] \\ =\frac{7\times 6\times 5\times 4!}{4!\left(3\times 2\times 1\right)} \\ =35 \end{gathered}$$

Therefore, option (C) is the correct answer.