The number of combinations with repetition that are allowed if n = 4 and r = 3 is given by: Option A: 20
The number of k-element combinations of n objects, with repetition is:
[tex]\overline{C}_{n,k} = \: ^{n + k-1}C_{k} = \dfrac{(n+k-1)!}{k! \times (n-1)!}[/tex]
For this case, we're given that:
Thus, we get:
[tex]\overline{C}_{n,k} = \dfrac{(n+k-1)!}{k! \times (n-1)!} = \dfrac{(4+3-1)!}{3! \times (4-1)!} \\\\\overline{C}_{n,k} = \dfrac{6!}{3!. 3!} = 20[/tex]
Thus, the number of combinations with repetition that are allowed if n = 4 and r = 3 is given by: Option A: 20
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