Mathematics Asked by user780357 on December 18, 2020

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!

- 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.
- 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).

Correct answer by Aniruddha Deb on December 18, 2020

For the first question: your attempt is related to the situation where the two groups are labeled. So, if the goal is to divide people equally into group A and B, then the answer is ${6 choose 3}$.

However, this question (seeing from the solution) does not presupposes that these group is labeled. In this case, having people 1,2,3 in group A and people 4,5,6 in group B is equivalent to 4,5,6 in A and 1,2,3 in B, for they divide these people in the same fashion. Because of this type of duality, you have to divide your answer by 2, the factorial of the cardinal of each cosets of this equivalent relation.

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