On Jan 27, 1:46 pm, "Brian M. Scott" <b.sc...@[EMAIL PROTECTED]
> wrote:
> On Sun, 27 Jan 2008 11:28:59 -0800 (PST), Tallison
> <Tallison.Rau...@[EMAIL PROTECTED]
> wrote in
> <news:037f548d-334f-43a5-9593-937005da0ff8@[EMAIL PROTECTED]
>
> in alt.algebra.help:
>
> > The questionut: How many five-card combinations of a standard playing
> > card deck have cards from exactly two suits?
> > The answer: 379,236.
> > The sad thing is that after working on this problem for the last hour,
> > I've gotten answers ranging from 32,712 to 65,780 to 268,382.4 (no
> > kidding!) to 328,900 - everything, that is, but the actual answer
> > according to the textbook.
> > So for the love of God, can someone help me identify the correct
> > formula?????
>
> There are C(4, 2) = 6 ways to choose the two suits to be
> represented in the hand. Now let A and B stand for any two
> of the suits. A 'legal' hand can have:
>
> (a) 1 card from A and 4 from B;
> (b) 2 cards from A and 3 from B;
> (c) 3 cards from A and 2 from B; or
> (d) 4 cards from A and 1 from B.
Too complicated. Just compute the number of ways to choose five cards
from those two suits and then subtract the hands where all the cards
are in the same suit:
C(26,5) - 2*C(13,5) = 65780 - 2574 = 63206
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