Question 1160504: Eight women and five men are standing in a line.
(a) how many arrangements are possible if any individual can stand in any position
(b) in how many arrangements will all five men be standing next to one another.
(c) in how many arrangements will no two men be standing next to one another.
Answer by ikleyn(52786)   (Show Source):
You can put this solution on YOUR website! .
It is strange to me to see placed into one capsule three totally different in their complexity subjects.
Part (a) is trivial and routine;
Part (b) is trivial and routine, too;
and finally part (c) is of the totally different level of complexity.
I will answer part (c), ONLY.
Imagine that the 8 women are standing in line with intervals (gaps) between them, as shown in the Figure below.
* W * W * W * W * W * W * W * W *
(" * "s show intervals, or gaps). You see 7 stars between "W"s and two stars before and after the W-line; in all, 7+2 = 9 stars.
These stars are the only places for men: Each man can occupy one and only one position at the star place.
So, we can position 8 women by 8! = 8*7*6*5*4*3*2*1 = 40320 ways.
And we can place first man in any of nine (*)-positions;
second man in any of remaining eight (*)-positions;
third man in any of remaining seven positions;
fourth . . .
and fifth . . .
Thus, in all, there are 8!*9*8*7*6*5 = 40320*9*8*7*6*5 = 609638400 differennt arrangements satisfying given restrictions. ANSWER
Solved.
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