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

Show that the number of permutation of n different things taken all at a time in which p particular thing are never together is
n!(np+1)!p!

Open in App
Solution

where p things are never together is
n!(np+1)!p!
When all the letters are taken, then they can be arranged in 10P10 i.e. 10 ways.
Take E, H, P as one letter and so the number of letters will be 103+1=8. As in part (d), the number of words in which E, H, P are together will be 8!.3!. The total number of arrangement by (a) is 10!. Hence the number of words when E, H, P are never together is
10!8!.3!=8![10×96]=84.8!.
Some students may try the question with gap method. Arrange the seven letters (E, H, P excluded) in 7! ways. We will have 8 gaps in which these three can be arranged in
8P3=8!5!=8.76ways.
hence the required number is
7!(8.76)=8!(76)=42.8!
This is half the number of ways we have calculated above.
Reason: Our condition is that the three letters E, H, P all are never together. Any two of them could be together, or all of them could be separated.

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