I only know the very basic formula for combinations: (n r)= n!/[r! (n-r)!]

I've looked at similar problems on this website, but I'm having trouble understanding how these problems are done exactly:

"In how many ways can 10 people be divided into two groups (one group of 6 and one group of four)?"

my sad attempt: (10 6) (4 4) = (10!/6!)*(4!/4!) = 5040???

I'd really appreciate it if someone could explain how to go about doing this step by step.


I will try to show to you, visually, a direct interpretation of the problem.

First of all you must notice that one of the main meaning of a factorial is the number of permutations of n different elements over n positions.

By example the number of permutations of the string $ABCDE$ is $5!=5\cdot 4\cdot 3\cdot 2\cdot 1$ (ways to order 5 different elements over 5 different positions).

A direct interpretation of your problem, trough permutations, is that you have a string of 10 different elements, e.g. $ABCDEFGHIJK$ or $0123456789$, and you divide this string in two groups: one of length $4$ (i.e. cardinality 4) and other of length $6$

$$\overbrace{ABCD}\ \overbrace{EFGHIJK}$$

You have $10!$ ways to order a string of 10 different elements, after you can divide each of these string on 2 groups of length $4$ and $6$. But you doenst care the order of each group, for your problem $AKJG=JAKG=GKAJ$. The number of ways to order a string of length $4$ is $4!$. The same for your string of length $6$, so you must eliminate all of these duplicates.

So you will have $\frac{10!}{4!\cdot 6!}$ ways to form these different groups of 4 and 6 people. For this problem, because the binomial coefficient is defined as $\binom{n}{k}=\frac{n!}{k!\cdot (n-k)!}$ then $\frac{10!}{4!\cdot 6!}=\binom{10}{4}$.

This is a direct interpretation of the problem using the meaning of factorial as number of permutations of n elements over n positions.

I will add an alternative, faster, way to interpret visually this problem.

First of all I will expand the meaning of factorial as the permutations of n different elements over n positions. If you have n elements (or objects) for the position one you can choose between n elements to put, for the second position you can choose between $n-1$ elements (because you have already one element on position 1). For position 3 you can choose between $n-2$ elements (because you have, already, some element on position 1 and 2), etc., etc., etc...

So the permutations will be $n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdots 1=n!$

Now our case: we have 10 people, and you want to see the different ways that you can group them in 6 and 4. The ordered ways for a group of 6 will be: for position 1 you can choose between 10, for position 2 between 9, ..., for position 6 between 5, i.e. $10\cdot9\cdot8\cdot7\cdot6\cdot5=(10)_6=\frac{10!}{4!}$ (the expression $(n)_k$ is named falling factorial).

But this is the expression of ordered positions. To have the unordered positions you must divide between the different ways to order 6 different elements on 6 positions, i.e. divide between $6!$. So the unordered ways to group 10 different elements in a group of 6 is $\frac{(10)_6}{6!}=\binom{10}{6}$.

But, what happen with the other group of length 4? For every group of length 6 we will have a unordered group of 4 so the total amounts to divide a group of 10 people in two groups of 6 and 4 is $\binom{10}{6}\cdot1=\binom{10}{6}$.


$$ \binom{10}{6} = \frac{10!}{6!\,4!} = \frac{10\cdot9\cdot8\cdot7\cdot\overbrace{6\cdot5\cdot4\cdot3\cdot2\cdot1}}{4\cdot3\cdot2\cdot1\cdot\underbrace{6\cdot5\cdot4\cdot3\cdot2\cdot1}} = \frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1} $$ Since $8$ cancels $4\cdot2$ and $\dfrac 9 3=3$, this reduces to $10\cdot3\cdot7$.

(If you wanted two groups of $5$ then you'd do a similar thing but you'd divide by $2$ when you're done, because one set of $5$ and its complementary set of the other $5$ would be the same way of dividing the set of $10$ into two sets of $5$. But that doesn't happen with $4$ and $6$ since $4\ne6$.)

  • $\begingroup$ Worth noting that it is equivalent to ${10 \choose 4}{6 \choose 6}$. $\endgroup$ – miniparser Nov 2 '14 at 20:01

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