Question

A. In how many ways can 5 children be arranged in a line such that
(1) two of them, Ram and Shyam, are always together,
(2) two of them, Ram and Shyam, are never together?​

Answer 1

Ram → R Shyam → S

No. of ways of arranging in which R & S together =4!×2!=48

No. of ways of arranging in which R & S not together = Total ways to arrange − No. of ways of arranging in which R & S together.

=5!−4!×2!

=120−48=72.