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There are 12 people in a dinner gathering. In how many ways can the host (one of the 12) arrange his guests around a dining table if a. they can sit on
any of the chairs? b. 3 people insist on sitting beside each other? c. 2 people refuse to sit beside each other?

Sagot :

Alphabetagammaidn
a. they can sit on any of the chairs
12! = 479,001,600 ways

b. 3 people insist on sitting beside each other
Count the 3 people as one, so the total number of people left is 10.
10! = 3,628,800
But the number of ways those 3 people can sit together is
3! = 6
So, if we get the total we simply multiply 10! and 3!
(10!)(3!) = (3,628,800)(6)
(10!)(3!) = 21,772,800 ways

c. 2 people refuse to sit beside each other
Let us count first the number of ways of those 2 people if they sit together. So, we count the 2 people as one, and the total number of people left is 11.
11! = 39,916,800
the number of ways the 2 can sit together is
2! = 2
Getting the total, we have
(11!)(2!) = 79,833,600
But the situation is, they refuse to sit beside each other. So, we subtract this answer to the total number of ways without restriction.
12! - (11!)(2!) = 479,001,600 - 79,833,600
12! - (11!)(2!) =  399,168,000 ways




- D.E.
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