# Math Help - combinations: possibly really easy?

1. ## combinations: possibly really easy?

how many different ways could you choose 6 cards from a standard deck of 52 cards, if you must have atleast one card from each suit, and order does not matter.

I got 8682544, is this correct?

2. Originally Posted by white_cap
how many different ways could you choose 6 cards from a standard deck of 52 cards, if you must have atleast one card from each suit, and order does not matter. I got 8682544, is this correct?
Yes that seems to be correct.
$\sum\limits_{k = 0}^3 {\left( { - 1} \right)^k \left( {\begin{array}{c}
4 \\
k \\
\end{array}} \right)\left( {\begin{array}{c}
{52 - 13 \cdot k} \\
6 \\
\end{array}} \right)}$

3. Hello, white_cap!

How many different ways could you choose 6 cards from a standard deck of 52 cards,
if you must have at least one card from each suit, and order does not matter.

I got 8682544, is this correct? . . . . I got the same answer!
I "talked" my way through it . . .

We choose six cards, and all four suits are represented.

There are two distributions of suits.

(1) There is 1 of the first suit, 1 of the second suit, 1 of the third suit,
. . and 3 of the fourth suit: . $[1,\,1,\,1,\,3]$

There are: .4 choices for the "triple" suit.
Then there are: . ${13\choose1}{13\choose1}{13\choose1}{13\choose3} \:=\:628,342$ ways to choose the cards.

Hence: there are: . $4 \times 628,342 \:=\:{\bf 2,513,368}$ ways to draw $[1,\,1,\,1,\,3]$

(2) There is 1 of the first suit, 1 of the second suit, 2 of the third suit,
. . and 2 of the fourth suit: . $[1,\,1,\,2,\,2]$

There are: . ${4\choose2,2} = 6$ to select the suits.
Then there are: . ${13\choose1}{13\choose1}{13\choose2}{13\choose2} \:=\:1,028,196$ ways to choose the cards.

Hence, there are: . $6 \times 1,028,196 \:=\:{\bf6,169,176}$ ways to draw $[1,\,1,\,2,\,2,]$

Therefore, there is a total of: . $2,513,368 + 6,169,176 \;=\;{\bf{\color{blue}8,682,544}}$ ways.