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I'm trying to get the hang of permutations and combinations for a discrete math class. One of my questions is as follows:

$$\text { You have a group of 30 people, and want to divide the group into two groups of 15 people each. }$$ $$\text { How many ways can this be done? } $$

My intuition is telling me to start with each group, say $G_{1}$ and $G_{2}$. Each group has the potential of selecting 30 people from the original group. Therefore, I'd calculate it as

$$\frac{30!}{(30-15)!}+\frac{30!}{(30-15)!}$$

But this doesn't seem right. If I'm done selecting all the possible people for $G_{1}$, then $G_{2}$ won't have $30!$ to choose from.

Could someone please clarify this for me?

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    $\begingroup$ If the groups are labelled it is $\binom{30}{15}$. If unlabelled, then $\frac{1}{2}\cdot \binom{30}{15}$. $\endgroup$ Oct 8, 2015 at 21:43

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This is a combination. To be precise, $_{30} C_{15}$. You can see this by realizing that having one group, say {1,2,3...15} implies that you are excluding half of the population, namely {16,17...30}. This isn't really the proper notation for counting these things, but it is appropriate nonetheless. It doesn't matter what the second group is, because they are being excluded because the other 15 were chosen.

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  • $\begingroup$ Your answer assumes the groups are labeled. If they are not labeled, you must multiply your result by $1/2$. $\endgroup$ Oct 9, 2015 at 9:58
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A first remark: it should be multiplication by the basic principle of counting, not addition, i.e, it should be $$\dfrac{30!}{15!}\times\dfrac{30!}{15!}$$

Now the way you count doesn't seem correct. In a group of people, we don't care about their order there, so you're counting each outcome in each group $15!$ more times. For example for the case of a group of $4$ people of $30$, if you pick $4$ people $P_1,P_2,P_3,P_4$ this is just the same as $P_3,P_2,P_4,P_1$ or any other permutation. Thus your formula should be $$\binom{30}{15}\times\binom{30}{15}$$ But wait! Once you select 15 people to form group 1, there are only 15 people left to choose for group two. So in fact the correct answer is:$$\binom{30}{15}\binom{15}{15}=\binom{30}{15}$$

Additional information: Your exercise is a specefic case of distributing $n$ objects into $k$ groups such as the $i$th group contains $n_i$ objects (so we must have $n_1+n_2+...+n_k=n$ (in your case $n=30$, $k=2$ and $n_1=n_2=15$) and there's a general formula for counting the number of such combinaisions using multinomial coefficients.

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  • $\begingroup$ Thank you @Scientifica, that was a really helpful explanation! $\endgroup$
    – thisisanon
    Oct 8, 2015 at 21:31
  • $\begingroup$ @ptikobj Glad to hear so :D $\endgroup$ Oct 8, 2015 at 22:06
  • $\begingroup$ Your answer assumes the groups are labeled. If they are not labeled, you must multiply your result by $1/2$. $\endgroup$ Oct 9, 2015 at 9:58
  • $\begingroup$ @N.F.Taussig You're right, but they're indeed labeled since that's what the OP assumed: $G_1$ and $G_2$. $\endgroup$ Oct 9, 2015 at 11:51

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