[Math] In how many ways can a group of six people be divided into: 2 equal groups? 2 unequal groups, if there must be at least one person in each group

combinatorics

In how many ways can a group of six people be divided into:

a) two equal groups

I have $^6C_3 \times \space ^3C_3 = 20$

So, to choose the first group I have $6$ possibilities of which I am choosing $3$. For the second group, I have $3$ remaining people of which $3$ must be chosen -> hence $^6C_3 \times ^3C_3 = 20$.

But the answer is $\frac{^6C_3}{2}$ but I don't understand why you divide by $2$.

b) two unequal groups, if there must be at least one person in each
group?

Applying the same logic as before, I got:

$$(^6C_2 \times ^4C_4) + (^6C_1 \times ^6C_5) = 51$$

But the answer is $^6C_1 + \space ^6C_2 = 21$

Could anyone explain how to solve these/the intuition behind it? Thanks in advance!

Best Answer

  1. If you're dividing 6 people into two groups $x$ and $y$, then $^6C_3$ would give you the total number of pairs of $(x,y)$ as well as $(y,x)$. To avoid the repetition of the pair $(y,x)$, you divide by two.
  2. The same logic applies here: when you split into groups, the remaining people whom you didn't select to form a group form the second group. To avoid repetition, the number of groups is $^6C_1 + \space ^6C_2 = 21$.

If you're still confused, the best way to grasp this concept is to take 4 people $a,b,c,d$ and see in how many ways you can split them into groups of 2 manually (by writing down all cases).

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