If eight persons are to address a meeting then the number of ways in which a specified speaker is to speak before another specified speaker, is
Answers
Answer:
20160
Step-by-step explanation:
If eight persons are to address a meeting then the number of ways in which a specified speaker is to speak before another specified speaker, is
Let say Two specified speakers are A & B
A has to speak before B
Rest 6 Speakers are others
Number of possible ways
if A speak 1st then B can speak from 2nd to 8th - 7 ways
and other 6 speaker can speak in 6! ways
= 7 * 6!
if A speak 2nd then B can speak from 3rd to 8th - 6 ways
and other 6 speaker can speak in 6! ways
= 6 * 6!
Similarly Number of position B can speak will keep reducing and other 6 speaker can speak in 6! ways
Total cases would be
7 * 6! + 6*6! + 5*6! + 4 *6! + 3*6! + 2*6! + 6! + 0
= 6!(7 + 6 + 5 + 4 + 3 + 2 + 1 )
=6! ( 28)
= 720 * 28
= 20160
Or Simply we can solve it other ways
8 speaker can speak in 8! ways
8! = 40320
Either one speaker will speak first oor other
so number of ways in which a specified speaker is to speak before another specified speaker, is = 40320/2 = 20160