Respuesta :
They can be arranged in 210 ways.
The arrangement of n objects where n₁ is one kind, n₂ is a second kind, and n₃ is a third kind is given by
[tex]\frac{n!}{n_1!\times n_2!\times n_3!}[/tex]
There are 7 total gems; 3 of the first kind, 2 of the second kind and 2 of the third kind:
[tex]\frac{7!}{3!2!2!} = 210[/tex]
The arrangement of n objects where n₁ is one kind, n₂ is a second kind, and n₃ is a third kind is given by
[tex]\frac{n!}{n_1!\times n_2!\times n_3!}[/tex]
There are 7 total gems; 3 of the first kind, 2 of the second kind and 2 of the third kind:
[tex]\frac{7!}{3!2!2!} = 210[/tex]
They can be arranged in 210 ways.
We have given that, 3 identical emeralds 2 identical diamonds and 2 different opals
What is the arrangement of n object where n₁ is one kind, n₂ is a second kind, and n₃ is a third kind?
The arrangement of n objects where n₁ is one kind, n₂ is a second kind, and n₃ is a third kind is given by
[tex]\frac{n!}{n_1!*n_2!*n_3!}[/tex]
The total number of gems =7
3 of the first kind, 2 of the second kind and 2 of the third kind
Therefore, we get
[tex]\frac{7!}{2!*2!*3!}=210[/tex]
Therefore ,they can be arranged in 210 ways.
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