Question: Eight persons amongst whom A, B and C are to speak at a function. Number of different ways can it do...

Question

Eight persons amongst whom A, B and C are to speak at a function. Number of different ways can it done if A wants to speak before B and B wants to speak before C is

Answer 1

Solution

Out of 8 persons we can select 3 persons in $8C_{3}$ ways, where A speaks before B and B speaks before C. A speaks before B and B speaks before C. Remaining 5 speakers can speak in 5! Ways.

∴ Total ways = $8C_{3}x5!$ = 6720