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
$n!-(n-p+1)!p!$

Answer 1

$\therefore$ where $p$ things are never together is
$n!-(n-p+1)!p!$
When all the letters are taken, then they can be arranged in ${}^{10}{P}_{10}$ i.e. $10$ ways.
Take E, H, P as one letter and so the number of letters will be $10-3+1=8$. 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
$10!-8!.3!=8![10\times 9-6]=84.8!$.
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
${}^8{P}_3=\frac{8!}{5!}=8.76$ways.
hence the required number is
$7!\left(8.76\right)=8!\left(76\right)=42.8!$
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.