From a standard 52-card deck, how many ways are there to pick a hand of k cards that includes one card from all four suits?I know that for any specific k, it's possible to break it up into cases based on the partitions of k into 4 parts. For example, if I want to choose a hand of six cards, I can break it up into two cases based on whether there are ( 1 ) three cards from one suit and one card from each of the other three or ( 2 ) two cards from each of two suits and one card from each of the other two.Is there a simpler, more general solution that doesn't require splitting the problem into many different cases?

Question

From a standard   52-card deck, how many ways are there to pick a hand of   k cards that includes one card from all four suits?I know that for any specific   k, it&#039;s possible to break it up into cases based on the partitions of   k into   4 parts. For example, if I want to choose a hand of six cards, I can break it up into two cases based on whether there are   (  1  ) three cards from one suit and one card from each of the other three or   (  2  ) two cards from each of two suits and one card from each of the other two.Is there a simpler, more general solution that doesn&#039;t require splitting the problem into many different cases?

Layla Love · Accepted Answer

Count the number of hands that do not contain at least one card from every suit and subtract from the total number of   k-card hands. To count the number of hands that do not contain at least one card from every suit, use inclusion-exclusion considering what suits are not in a given hand. That is, letting   N  (  …  ) mean the number of hands meeting the given criteria,                          N        (                  n          o                           ♡        )        +        N        (                  n          o                           ♠        )        +        N        (                  n          o                           ♣        )        +        N        (                  n          o                           ♢        )                                                          −        N        (                  n          o                           ♡        ♠        )        −        N        (                  n          o                           ♡        ♣        )        −        N        (                  n          o                           ♡        ♢        )        −        N        (                  n          o                           ♠        ♣        )        −        N        (                  n          o                           ♠        ♢        )        −        N        (                  n          o                           ♣        ♢        )                                                          +        N        (                  n          o                           ♡        ♠        ♣        )        +        N        (                  n          o                           ♡        ♠        ♢        )        +        N        (                  n          o                           ♡        ♣        ♢        )        +        N        (                  n          o                           ♠        ♣        ♢        )                                                          −        N        (                  n          o                           ♡        ♠        ♣        ♢        )                                          =        4                                              (                                39            k                                          )                          −        6                                              (                                26            k                                          )                          +        4                                              (                                13            k                                          )                          −                                              (                                0            k                                          )                          .            So, the number of hands of   k cards that include at least one card from every suit is                (              52      k                  )        −  4                (              39      k                  )        +  6                (              26      k                  )        −  4                (              13      k                  )        +                (              0      k                  )        .