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

There are 5 different boxes and 7 different balls. All the 7 balls are to be distributed among the 5 boxes placed in a row so that any box can receive any number of balls.

In how many ways can these balls be distributed such that no box is empty and ball 2 and ball 4 cannot be put in the same box?

A
1200
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
15000
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3800
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 15000
The required number of distribution = Total number of distribution - Number of distribution in which ball 2 and ball 4 are together.
Here the total number of distributions = 16800.
Now, consider ball 2 and ball 4 are stuck together and this arrangement is assumed to be a single ball.
Thus we have 6 balls to be distributed into 5 boxes.Which can be done as 1, 1, 1, 1, 2 in 5C1×6C2×4! ways=1800.
5C1 Number of ways of selecting one box in which two balls are kept together.
6C2 Number of ways of selecting 2 balls out of 6 balls.
4! Number of ways of distribution of remaining 4 balls in remaining 4 boxes.
Hence, the required number of ways = 16800 - 1800 = 15000

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Different to Different
QUANTITATIVE APTITUDE
Watch in App
Join BYJU'S Learning Program
CrossIcon