wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The total number of ways in which six + and four signs can be arranged in a line such that no two signs occur together is ______.

Open in App
Solution

+ signs can be put in a row in one way creating seven gaps shown as arrows:
Now 4 signs must be kept in these gaps. So, no two signs should be together.
Out of these 7 gaps 4 can be chosen in 7C4 ways. Hence, required number of arrangements is
7C4=7C3=7×6×53×2×1=35.
1569055_145674_ans_46501f63d1314d05a1a499c789368078.jpg

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon