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?
Hello, white_cap!
I "talked" my way through it . . .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!
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: .$\displaystyle [1,\,1,\,1,\,3]$
There are: .4 choices for the "triple" suit.
Then there are: .$\displaystyle {13\choose1}{13\choose1}{13\choose1}{13\choose3} \:=\:628,342$ ways to choose the cards.
Hence: there are: .$\displaystyle 4 \times 628,342 \:=\:{\bf 2,513,368}$ ways to draw $\displaystyle [1,\,1,\,1,\,3]$
(2) There is 1 of the first suit, 1 of the second suit, 2 of the third suit,
. . and 2 of the fourth suit: .$\displaystyle [1,\,1,\,2,\,2]$
There are: .$\displaystyle {4\choose2,2} = 6$ to select the suits.
Then there are: .$\displaystyle {13\choose1}{13\choose1}{13\choose2}{13\choose2} \:=\:1,028,196$ ways to choose the cards.
Hence, there are: .$\displaystyle 6 \times 1,028,196 \:=\:{\bf6,169,176}$ ways to draw $\displaystyle [1,\,1,\,2,\,2,]$
Therefore, there is a total of: .$\displaystyle 2,513,368 + 6,169,176 \;=\;{\bf{\color{blue}8,682,544}}$ ways.