Answer:
15 different combinations
Step-by-step explanation:
Sara has six necklaces, but her mother will allow her to wear two at a time.
So we calculate different combinations of two necklaces by this formula:
[tex]\frac{n!}{r!(n-r)!}[/tex]
Where n = 6 and r = 2
Now put the values
[tex]\frac{6!}{2!(6-2)!}[/tex]
[tex]\frac{6\times 5\times 4\times 3\times 2\times 1}{2!(6-2)!}[/tex]
[tex]\frac{720}{2\times 1(4)!}[/tex]
[tex]\frac{720}{2\times 1(4\times 3\times 2\times 1)}[/tex]
[tex]\frac{720}{2(24)}[/tex]
[tex]\frac{720}{48}[/tex] = 15 combinations
Sara can wear 15 different combinations of two necklaces.
