A question about the relation between the combination and the type of objects: Should the objects be incongruous?

Question

I'd been taught that the number of ways that we can select $k$ objects from $n$ nonidentical objects, is given by (regardless of order):

$$C_{n}^{k}=\frac{n!}{k!*(n-k)!}$$

I have a very simple question about this matter:

Is it really obligatory that the objects are nonidentical?

If so, why in the ball-examples we use combination? For instance, consider we have $4$ red balls and $3$ green balls in a container. Then, we are asked to say the number of ways that we can select two balls from the container. As you know, we immediately say that the number of ways that we can choose $2$ balls from the container is $C_{7}^{2}$. However, all the balls are not nonidentical. With this regard, what would be the number of ways that we can choose $2$ red balls from $6$ red balls? Is it $C_{2}^{6}$?

Answer 1

Yes, it only works if the items are non-indentical.   $^n\mathrm C_r$ counts the distinct ways to select $r$ elements from a set of $n$ indistinguishable elements.   If the elements were identical, there would only be one distinct way to select $r$ of them.   (Assuming integer values and $0\leq r\leq n$.)

Note: Physically the balls will be unique entities, but what we are counting is the arrangements of identifiable traits.   (Such as colours.)

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PS: ${^n\mathrm C_r} = \frac{n!}{r!(n-r)!}$.   I've encountered it written as $\mathrm C^n_r$, but always with the superscript corresponding to the numerator while the subscript to the denominator.   It would be confusing for a text to present it otherwise.

Answer 2

Working on the case that you sugested, you have three ways to choose those balls:

$$(red,red), (red,green), (green,green)$$

You will always fall in one of those cases. So, to be non-identical really matters.

Like @Graham said:

Physically the balls will be unique entities, but what we are counting is the arrangements of identifiable traits. (Such as colours.)