How many ways can a hand of five cards consisting of three cards from one suit and two cards from another suit be drawn from a standard deck of cards?

Given to draw a hand of five cards consisting of three cards from one suit and two cards from another suit from a standard deck of cards. The number of ways the hand can be drawn is to be determined.

A permutation is when n objects are available and r are to be picked and arranged in a certain order and the number of permutations is given by $P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$

A combination is when n objects are available and r are to be picked without arrangement and the number of combinations is given by $C\left(n,r\right)=\frac{n!}{\left(n-r\right)!r!}$

Here, first two of the four suits is to be arranged as one suit gives four cards and another suit gives one card which can be done in $P\left(4,2\right)$ ways.

Three cards of the thirteen cards in a suit can be selected in $C\left(13,3\right)$ ways.

Two cards in the second suit can be selected in $C\left(13,2\right)$ ways.

Hence the total number of ways is given by:

$\begin{array}{l}P\left(4,2\right)\cdot C\left(13,3\right)\cdot C\left(13,2\right)=\frac{4!}{\left(4-2\right)!}\cdot \frac{13!}{\left(13-3\right)!3!}\cdot \frac{13!}{\left(13-2\right)!2!}\\ P\left(4,2\right)\cdot C\left(13,3\right)\cdot C\left(13,2\right)=\frac{4!}{2!}\cdot \frac{13!}{10!3!}\cdot \frac{13!}{11!2!}\\ P\left(4,2\right)\cdot C\left(13,3\right)\cdot C\left(13,2\right)=\frac{4\cdot 3!}{2\cdot 1}\cdot \frac{13\cdot 12\cdot 11\cdot 10!}{10!3!}\cdot \frac{13\cdot 12\cdot 11!}{11!\cdot 2\cdot 1}\\ P\left(4,2\right)\cdot C\left(13,3\right)\cdot C\left(13,2\right)=13\cdot 12\cdot 11\cdot 13\cdot 12\\ P\left(4,2\right)\cdot C\left(13,3\right)\cdot C\left(13,2\right)=267696\end{array}$

Hence, the number of ways the hand can be drawn is 267696.