Question

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

Answer 1

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