Six ‘ + ’ and four ‘ - ’ signs are to placed in a straight line so that no t

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6.Permutation and Combination

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Six ‘$+$’ and four ‘$-$’ signs are to 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

$42$

(IIT-1988)

Solution

(c) The arrangement can be make as $.+.+.+.+.+.+.$, the $( – )$ signs can be put in $7$ vacant (pointed) place.

Hence required number of ways ${ = ^7}{C_4} = 35$.

Standard 11

Mathematics

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