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

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

Vedclass · Accepted Answer

$35$

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(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 need to choose $4$ gaps out of these $7$ available positions to place the $4$ '$-$' signs.The number of ways to do this is given by the combination formula ${^n}C_r = \frac{n!}{r!(n-r)!}$.Here,$n = 7$ and $r = 4$,so the 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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