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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$35$

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 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:

Similar Questions

In how many ways can a student choose a programme of $5$ courses if $9$ courses are available and $2$ specific courses are compulsory for every student?

There are $7$ identical white balls and $3$ identical black balls. The number of distinguishable arrangements in a row of all the balls,so that no two black balls are adjacent is

Out of $10$ white,$9$ black and $7$ red balls,the number of ways in which selection of one or more balls can be made,is

If $^nC_{r-2} = 36$,$^nC_{r-1} = 84$,and $^nC_r = 126$,then the value of $^nC_{2r}$ is

From a class of $25$ students,$10$ are to be chosen for an excursion party. There are $3$ students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Vedclass Test Series

Exam Paper Generator

Online Exam Module