Six '+' and four '-' signs are to be placed i | vedclass

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Question

(C) To ensure that no two '$-$' signs come together,we first arrange the six '$+$' signs in a row.This creates $7$ possible gaps (including the ends)

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(C) To ensure that no two '$-$' signs come together,we first arrange the six '$+$' signs in a row.This creates $7$ possible gaps (including the ends) where the '$-$' signs can be placed: $\_ + \_ + \_ + \_ + \_ + \_ \_$.We have $4$ '$-$' signs to place in these $7$ available gaps.The number of ways to choose $4$ gaps out of $7$ is given by the combination formula ${^n}C_r = \frac{n!}{r!(n-r)!}$.Therefore,the total number of ways is ${^7}C_4 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.

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 is:

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