If there are 4 chairs and 5 people, how many possible combinations of seating arrangements are there? In this case, the order of people in the chairs does not matter.
If there are 4 chairs and 5 people, how many possible permutations of seating arrangements are there? In this case, the order of people in the chairs does matter.
Solution :
1)
n C x = n! / x!(n - x)!
n! = n(n - 1)(n - 2) .... and so on .
n = 5 and x = 4
5 C 4 = 5! / 4!(1)! = 5
2)
n P x = n! / (n - x)!
n! = n(n - 1)(n - 2) .... and so on .
n = 5 and x = 4
5 P 4 = 5! / (1)! = 120
If there are 4 chairs and 5 people, how many possible combinations of seating arrangements are...
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