Another combinatorics problem involving permutations and counting methods

Question

A course with 2 of 'a' subjects, 3 of 'b' subjects and 2 of 'c' subjects is to be ordered in a row. The number of arrangments with 'b' subjects together is?

Again, I'm not sure what I've left out here as shouldn't the answer just be $^3 P_3$ ?

Answer 1

No, the answer is not $^{3}P_3=\frac{3!}{0!}=6$.

You should see the group of the three 'b' subjects as one object, because they must be together.

So you want to order two 'a' subjects, a 'bbb' subject and two 'c' subjects.
Do you know what the answer is now?

Answer 2

If:

• $\color\red {a}$ subjects are identical to each other
• $\color\green {b}$ subjects are identical to each other
• $\color\orange{c}$ subjects are identical to each other

Then the answer is $\frac{(\color\red2+\color\green1+\color\orange2)!}{\color\red2!\times\color\green1!\times\color\orange2!}=30$

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If:

• $\color\red {a}$ subjects are not identical to each other
• $\color\green {b}$ subjects are not identical to each other
• $\color\orange{c}$ subjects are not identical to each other

Then the answer is $(\color\red2+\color\green1+\color\orange2)!\times\color\green3!=720$