Answer:- 1
Explanation:-
Solution:
5 students can be seated out of 10 students in 10C5 ways . Remaining 5 will be seated in,= 5C5 ways .
Students of each row can be arranged as, = 5! *5! ways. Two sets of paper can be arranged themselves in, = 2! ways. Thus, Total arrangement,= 10C5 * 5! *5! *2
= 7257600.
in a row, means there are not seating in a circular seats
but instead, on a straight row of seats
if two students insists to sit beside each other
therefore, the two seat they will occupied will be count as one
so, we have 5 - 1 = 4 seats
P(10, 4)
= 10!/(10 - 4)!
= 10!/6!
= 5040 ways
now, we can arrange the two who insists on sitting together.
2! = 2
multiply the answers
5040*2
= 10,080 ways
-
120 ways is the correct answer
$\begingroup$
What I'm thinking: Find total ways that the ten people can be seated, which is 10!.
Then I take that number and subtract the ways the these two people would be seated next to each other. I do this by treating these two people as a single space, which leaves the eight other students plus that space consisting of the two. This would mean 9!
Then, 10! - 9! = 3265920 ways for the ten people to be seated so that a certain to are not next to each other.
$\endgroup$ 2