On Jan 27, 2:46=A0pm, "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: =A0How many five-card combinations of a standard
playing=
> > card deck have cards from exactly two suits?
> > The answer: =A0379,236.
> > 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) =3D 6 ways to choose the two suits to be
> represented in the hand. =A0Now let A and B stand for any two
> of the suits. =A0A 'legal' hand can have:
>
> =A0 =A0(a) =A01 card from A and 4 from B;
> =A0 =A0(b) =A02 cards from A and 3 from B;
> =A0 =A0(c) =A03 cards from A and 2 from B; or
> =A0 =A0(d) =A04 cards from A and 1 from B.
>
> There are C(13, k) ways to choose k cards from a single
> suit, and the choices from suits A and B are independent.
> Thus, there are
>
> =A0 =A0C(13, 1) * C(13, 4) =3D
> =A0 =A013 * (13*12*11*10)/(4*3*2*1) =3D
> =A0 =A013 * 715 =3D 9295
>
> hands in each of cases (a) and (d), and
>
> =A0 =A0C(13, 2) * C(13, 3) =3D
> =A0 =A0[(13*12)/(2*1)] * [(13*12*11)/(3*2*1)] =3D
> =A0 =A078 * 286 =3D 22,308
>
> hands in each of cases (b) and (c). =A0The grand total is
> therefore
>
> =A0 =A06 * (2 * 9295 + 2 * 22,308) =3D 6 * 63,206 =3D
> =A0 =A0379,236.
>
> Brian
God, Brian, thank you.
What I was missing was the C(6,2) ways by which I needed to multiply
the "a" through "d" product sums.
Great exposition! I've had about ten friends and colleagues working
on this problem (including physicians, as I'm an RN!) and you're the
first to get it.
Your explanation is so clear; you must be a teacher/instructor?
Thanks again! -Tallison
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