In how many ways can 6 '+' signs and 4 '*' si | vedclass

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(A) First,we arrange the $6$ '$+$' signs in a row: $+ + + + + +$.There are $7$ possible gaps (including the ends) created by these $6$ '$+$' signs: $\

Vedclass · Accepted Answer

$35$

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(A) First,we arrange the $6$ '$+$' signs in a row: $+ + + + + +$.There are $7$ possible gaps (including the ends) created by these $6$ '$+$' signs: $\_ + \_ + \_ + \_ + \_ + \_ \_$.To ensure no two '$*$' signs occur together,we must place the $4$ '$*$' signs into these $7$ available gaps.The number of ways to choose $4$ gaps out of $7$ is given by the combination formula $^nC_r = \binom{n}{r}$.Therefore,the total number of ways is $^7C_4 = \frac{7 	imes 6 	imes 5 	imes 4}{4 	imes 3 	imes 2 	imes 1} = 35$.

In how many ways can $6$ '$+$' signs and $4$ '$$' signs be arranged in a row such that no two '$$' signs occur together?

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In how many ways can $6$ '$+$' signs and $4$ '$*$' signs be arranged in a row such that no two '$*$' signs occur together?