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I am trying to determine the number of permutations of three children in a family, considering their gender. When listing out the permutations, I found eight possibilities: BBB, BBG, BGB, GBB, GGG, GGB, BGB, GBB. This can be calculated using the Fundamental Theorem of Counting ($2^3$). However, when I attempted to find it using $nPr$ I encountered a discrepancy.

The formula for permutations of $r$ objects taken at a time from n objects, where there are, say, $p$ objects of one type, $q$ objects of another type, and so on, is given as $\large{\frac{nPr}{ p!\times q!...}}$. This is also known as permutation with repetition formula.

In this children problem, we can consider it as selecting 3 children from a set of 6 children, with 3 being boys and 3 being girls. Thus, we have $p = q = 3$. Therefore, the formula should be $\large{\frac{6P3}{ 3! \times 3!}}$, which however, equals $\large\frac{10}{3}$. As you can see, this result is not consistent with the expected outcome of 8 permutations.

I seek clarification on how to properly use the $nPr$ formula in this context to obtain the correct number of permutations, which should be $8$.

Osmium
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  • Suppose your first child will be named Alice or Adrian, your second Bob or Barbara, and your third Carol or Collin. You cannot have both Alice and Adrian - that is, you cannot have two first children, only one first child - and similarly for Bob/Barbara and Carol/Collin. What you are supposed to be counting is not called a permutation. (Your formula isn't even right anyway: combinations are $nCr=nPr/r!$, so you divided by $3!$ one too many times for combinations.) – coiso Oct 16 '23 at 14:57
  • @coiso But BBG, BGB and GBB are different in the problem. So, we are dealing with permutations and not combinations. Right? – Osmium Oct 16 '23 at 15:10
  • While you used the word "permutation" and the wrong formula for combinations, what you're dealing with is neither. Your 3-out-of-6 interpretation is wrong, for the reason I gave in my previous comment. – coiso Oct 16 '23 at 15:31

1 Answers1

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"The formula for permutations of $r$ objects taken at a time from n objects, where there are, say, $p$ objects of one type, $q$ objects of another type, and so on, is given as $\large{\frac{nPr}{ p!\times q!...}}$. This is also known as permutation with repetition formula."

This is wrong, and is where you erred. Here is the correct version:

Permutations with repetition. Suppose there are $n$ objects total, in $k$ types, such that there are $n_i$ identical objects of the $i^\text{th}$ type for each $i \in \{1,\dots,k\}$. This means $n=n_1+\dots+n_k$. The number of ways to order all $n$ of the objects in a line is$$\frac{n!}{n_1!n_2!\cdots n_k!}$$

Notice that you cannot use this formula when you are doing "partial permutations", i.e. when you are only lining up some of the objects. Here is an MSE question which deals with partial permutations with repition: How to find the number of $k$-permutations of $n$ objects with $x$ types, and $r_1, r_2, r_3, \cdots , r_x$ = the number of each type of object?. None of the answers give a nice closed formula, because none exists.

For your problem, you can alternatively compute the number of partial permutations of the multiset $\{B,B,B,G,G,G\}$ of length $3$ as follows. First, select three objects from this multiset without caring about order. There are four possibilities: $BBB,BBG,BGG$ and $GGG$. For each of these possibilities, we then count the number of ways to order them using the permutations with repetition formula. The result is $$ \underbrace{\frac{3!}{3!}}_{BBB} +\underbrace{\frac{3!}{2!\cdot 1!}}_{BBG} +\underbrace{\frac{3!}{1!\cdot 2!}}_{BGG} +\underbrace{\frac{3!}{3!}}_{GGG}=8, $$ which agrees with $2^3$.

Mike Earnest
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  • Thank you very much for your nice answer. – Osmium Oct 17 '23 at 00:43
  • I have this one question more: is there a formula for "combination with repetition"? I didn't find it my book. I mean how can we find the four possibilities: BBB,BBG,BGG and GGG with a formula? – Osmium Oct 17 '23 at 01:16
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    For combinations with repetition: there are $n$ objects in $k$ types, with $n_1,\dots,n_k$ identical objects in each type. We want to choose a combination of $m$ objects. All that matters is how many objects chosen are from each type. Suppose there are $x_i$ items from $i^\text{th}$ type. Then $x_1+\dots+x_k=m$, and $0\le x_i\le n_i$, so we need to count the number of integer solutions. Finally, the methods for counting solutions are described at this question: https://math.stackexchange.com/questions/553960/extended-stars-and-bars-problemwhere-the-upper-limit-of-the-variable-is-bounded – Mike Earnest Oct 17 '23 at 01:42