How many permutations can be formed from $2n$ distinguishable objects and $n$ indistinguishable objects?

Please tell me if I am on the right track to solving this question.

Basic Formula: $C(n+r-1,n-1)$.

$r = n$

$n = 2n$

$C((2n+n-1),(2n-1))$ so that is equal to $\dfrac{(3n-1)!}{(2n-1)!(n)!}$.

$\dfrac{(3n-1)!}{(2n-1)!(n)!}$ permutations can be formed.


If we wish to permute the letters $ABCDEFGGG$, we can choose three of the nine locations for the $G$'s. Once they have been placed, we can place the remaining six letters in $6!$ orders, giving us $$\binom{9}{3} \cdot 6! = \frac{9!}{6!3!} \cdot 6! = \frac{9!}{3!}$$ distinguishable permutations.

In general, if we have $2n$ distinguishable objects and $n$ indistinguishable objects, we can choose $n$ of the $3n$ positions for the indistinguishable objects, then place the remaining $2n$ objects in $(2n)!$ ways. Hence, the number of ways of arranging $2n$ distinguishable objects and $n$ indistinguishable objects is $$\binom{3n}{n} \cdot (2n)! = \frac{(3n)!}{n!(2n)!} \cdot (2n)! = \frac{(3n)!}{n!}$$

Note: The formula $\binom{n + r - 1}{n - 1}$ is the number of ways of selecting $r$ objects of $n$ different types. For instance, it is used when you want to calculate the number of ways of selecting $12$ donuts if six varieties of donuts are available.


You can look at any problem of this type with repetition of objects, e.g. $AAABBCDDDDE,$

as firstly permuting the total number of objects, and then removing the permutations of objects that are identical, thus $\dfrac{11!}{3!2!4!}$

You should remember this as a general formula:
If there are $n$ objects with $n_1$ of one type, $n_2$ of another and so on,
number of distinct permutations = $\dfrac{n!}{n_1!n_2!....}$


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