# The number of ways of dividing 52 cards amongst four players so that three players have 17 cards each and the fourth player is left with only one card is?

$\frac{\left(\begin{matrix}52 \\ 17\end{matrix}\right) \left(\begin{matrix}35 \\ 17\end{matrix}\right) \left(\begin{matrix}18 \\ 17\end{matrix}\right) \left(\begin{matrix}1 \\ 1\end{matrix}\right)}{6} \approx 2.99 \times {10}^{23}$ ways

#### Explanation:

Let's first see that this is a combinations problem - we don't care about the order in which the cards are dealt:

C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

One way we can do this is to see that for the first person, we'll be choosing 17 from 52 cards:

$\left(\begin{matrix}52 \\ 17\end{matrix}\right)$

For the second person, we'll be picking 17 cards from the remaining 35 cards:

$\left(\begin{matrix}52 \\ 17\end{matrix}\right) \left(\begin{matrix}37 \\ 17\end{matrix}\right)$

and we can do the same thing for the next player:

$\left(\begin{matrix}52 \\ 17\end{matrix}\right) \left(\begin{matrix}35 \\ 17\end{matrix}\right) \left(\begin{matrix}18 \\ 17\end{matrix}\right)$

and we can enter a last term for the last player as well:

$\left(\begin{matrix}52 \\ 17\end{matrix}\right) \left(\begin{matrix}35 \\ 17\end{matrix}\right) \left(\begin{matrix}18 \\ 17\end{matrix}\right) \left(\begin{matrix}1 \\ 1\end{matrix}\right)$

And now for the last bit - we've set this up so that there is a definite first person, then second person, then third person, then last person - which could be ok but we're treating the first person differently than the second and those two are different from the third, even though they are supposed to be identical in their drawing method. We've made order important and order is a permutation concept (see below for more on this).

We don't want to have the order be important and so need to divide by the number of ways we can arrange the three people - which is 3! = 6

This all gives:

$\frac{\left(\begin{matrix}52 \\ 17\end{matrix}\right) \left(\begin{matrix}35 \\ 17\end{matrix}\right) \left(\begin{matrix}18 \\ 17\end{matrix}\right) \left(\begin{matrix}1 \\ 1\end{matrix}\right)}{6} \approx 2.99 \times {10}^{23}$ ways

~~~~~

Let's look at a much smaller example to see the note on order. Let's take 5 items and distribute them among 3 people: 2 people get 2 items each and the last person gets the remaining item. Calculating the same way we did above:

$\left(\begin{matrix}5 \\ 2\end{matrix}\right) \left(\begin{matrix}3 \\ 2\end{matrix}\right) \left(\begin{matrix}1 \\ 1\end{matrix}\right) = 10 \times 3 \times 1 = 30$ ways

But if we actually count them out:

A, BC, DE
A, BD, CE
A, BE, CD

B, AC, DE
B, AE, CD

C, AB, DE
C, AE, BD

D, AB, CE
D, AC, BE
D, AE, BC

E, AB, CD
E, AC, BD
there are only 15. Why? We made a definite first person and second person in the calculation (one gets to pick from 5, the next to pick from 3) and so we made order matter. By dividing by the number of people who are supposed to be equal but are not in the calculation, we divide out the order, or the number of people who are supposed to be equal but aren't, factorial. In this case, that number is 2 and so 2! = 2, giving:
$\frac{30}{2} = 15$ which is the correct answer