On Sun, 27 Jan 2008 11:28:59 -0800 (PST), Tallison
<Tallison.Rausch@[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.
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
C(13, 1) * C(13, 4) =
13 * (13*12*11*10)/(4*3*2*1) =
13 * 715 = 9295
hands in each of cases (a) and (d), and
C(13, 2) * C(13, 3) =
[(13*12)/(2*1)] * [(13*12*11)/(3*2*1)] =
78 * 286 = 22,308
hands in each of cases (b) and (c). The grand total is
therefore
6 * (2 * 9295 + 2 * 22,308) = 6 * 63,206 =
379,236.
Brian
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