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?

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- Oct 22nd 2007, 08:28 PMwhite_capcombinations: 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? - Oct 23rd 2007, 03:56 AMPlato
- Oct 23rd 2007, 04:43 PMSoroban
Hello, white_cap!

Quote:

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.