We can solve this problem using fundamental principle of counting.
Fundamental Principle of Multiplication:
If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then two jobs in succession can be completed in m × n ways.
By using this definition we can find the answer.
Let the 7 people be A, B, C, D, E, F and G. There are 7 positions to be filled D, E, F and G can occupy any of the 7 positions.
Thus :
Number of options for D = 7
(Any of the 7 position)
Number of options for E = 6
(Any of the 6 remaining positions)
Number of options for F = 5
(Any of the 5 remaining positions)
Number of options for G = 4
(Any of the 4 remaining positions)
Since, A must speak before B and B must speak before C, the 3 remaining positions must be occupied as follows :
A - B - C
Thus,
number of options for A = 1
(A must occupy the leftmost remaining position)
Number of options for C = 1
(C must occupy the right most remaining positions)
Number of options for B = 1
(Only 1 position left)
To combine all of these options, we multiply :
7 × 6 × 5 × 4 × 1 × 1 × 1 = 840
Hence, the number of ways in which 7 persons can address a meeting so that out of the three persons A, B and C, A will speak before B and B before C, is 840.
Another method :
The same problem can be done by permutation also. If you do it by permutation we get
We have
Let the 7 people be A, B, C, E, F and G. Out of the three persons A, B and C. A will speak before B and B before C.
Then arrange the remaining 4 peoples in 7 places. This is done in
P4 ways. Then arrange ABC is
P4 = 7 × 6 × 5 × 4 × 1 × 1 × 1 = 840