On Feb 3, 1:06 am, "Brian M. Scott" <b.sc...@[EMAIL PROTECTED]
> wrote:
> On Sat, 2 Feb 2008 22:50:31 -0800 (PST), kevin cline
> <kevin.cl...@[EMAIL PROTECTED]
> wrote in
> <news:9af82bfd-5b0b-416d-a27c-d3e7d1f1a959@[EMAIL PROTECTED]
>
> in alt.algebra.help:
>
>
>
> > 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
>
> In this problem the two approaches are of roughly comparable
> difficulty, so the one to be preferred when solving the
> problem is whichever one you see first. In this case I'd
> say that it's pretty much a toss-up for paedagogical
> purposes, too, though I lean towards the more
> straightforward analysis -- mine -- for beginners.
Great. Now use your method to compute the number of two-suited bridge
hands of 13 cards.
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