Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

In how many ways can 7 plus (+) signs and 4 minus (-) signs can be arranged in a row so that no two minus signs are together?

$\begin{array}{1 1}(11)! \\ ^7C_4 \\ 7!.4! \\ 70 \end{array}$

Can you answer this question?

1 Answer

0 votes
Since there is no condition for plus (+) sign, fix them in a row.
+ + + + + + +.
There are 6 places in between each plus and one before and one after these plus (+) signs.
$i.e.,$ there are 8 places for minus (-) and 4 (-) signs are there.
$\therefore$ these minus (-) signs can be placed in $^8C_4$ ways.
The required no. of arrangements = $^8C_4=70$


answered Aug 20, 2013 by rvidyagovindarajan_1
edited Dec 20, 2013 by meenakshi.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App