Recall that $P(n, k)$ is the symbol for the number of permutations of $n$ objects taken $k$ at a time and $C(n, k)=\left(\begin{array}{l}n \\ k\end{array}\right)$ is the symbol for the number of combinations of $n$ objects taken $k$ at a time. Using this notation we see that the number of different bridge hands is $C(52,13)=52 ! /(13 !)(39 !)=6.350135 \times 10^{11}$. We can compute the probability of a given distribution of cards by suit (such as 12 in one suit and one in another) in a randomly chosen hand by dividing the number of possible hands with such a distribution by $C(52,13)$. Consider a 5-4-3-1 distribution. If the suits are given (say the fivecard suit is hearts, the four-card suit diamonds, the three-card suit clubs, and the remaining card is a spade), there are
$$
\begin{aligned}
C(13,5) C(13,4) C(13,3) C(13,1) & =1,287 \times 715 \times 286 \times 13 \\
& =3,421,322,190
\end{aligned}
$$
such hands. But there are $P(4,4)=24$ ways of permuting the 4 different sized suits in a 5-4-3-1 distribution so we have
$$
\begin{aligned}
P[5-4-3-1 \text { distribution }] & =\frac{P(4,4) C(13,5) C(13,4) C(13,3) C(13,1)}{C(52,13)} \\
& =0.129307054 .
\end{aligned}
$$
(a) Show that the probability of a 4-4-3-2 distribution is
$$
\begin{aligned}
P[4-4-3-2 \text { distribution }] & =\frac{P(4,2) C(13,4)^2 C(13,3) C(13,2)}{C(52,13)} \\
& =0.215511757 .
\end{aligned}
$$
(b) Show that the probability of a 4-3-3-3 distribution is
$$
\begin{aligned}
P[4-3-3-3 \text { distribution }] & =\frac{P(4,1) C(13,4) C(13,3)^3}{P(52,13)} \\
& =0.105361303 .
\end{aligned}
$$
(c) Show that for any specific $\mathrm{x}-\mathrm{y}-\mathrm{z}-\mathrm{w}$ distribution (where $x+y+$ $z+w=13$ ), we have
$$
\begin{gathered}
P[\mathrm{x}-\mathrm{y}-\mathrm{z}-\mathrm{w} \text { distribution }]=\frac{n C(13, x) C(13, y) C(13, z) C(13, w)}{C(52,13)} \\
n=\left\{\begin{array}{l}
P(4,4)=24 \text { if all suits are of different size } \\
P(4,2)=12 \text { if exactly } 2 \text { suits are of the same size } \\
P(4,1)=4 \text { if } 3 \text { suits are of the same size. }
\end{array}\right.
\end{gathered}
$$
Of course, $n$ is the number of different suit arrangements for a given $\mathrm{x}-\mathrm{y}-\mathrm{z}-\mathrm{w}$ distribution.