Cards are drawn one by one in a pack of well shuffled cards without  replacement until all the cards are drawn then

Let us make a sample space of 52 dimensions (I.e.,) each point of sample in an ordered point of the form $\left\{{\mathrm{x}}_1,{\mathrm{x}}_2\dots \dots {\mathrm{x}}_{52}\right\}$
Total number of sample point are 52! 
Sample point on which x₁₃ is spade is $13{\mathrm{C}}_1\times 5!$
Probability of geeting spade on ${\mathrm{x}}_{13}=\frac{{ }^{13}{\mathrm{C}}_1\times 51!}{52!}=\frac{1}{4}$
Similarly number of sample points on x₅₂ is spade ${=}^{12}{\mathrm{C}}_1\times 51!$
Number of sample points having king at x₅ and queen at x₁₀ will be ${ }^4{\mathrm{C}}_1\times 50 !$
Required probability$=\frac{16\times 50!}{52!}=\frac{4}{663}$
Passage I:
Let $\mathrm{A}=\left\{{\mathrm{x}}_1,{\mathrm{x}}_2\dots \dots {\mathrm{x}}_{10}\right\}$  then for I=1 to 10 
$\begin{array}{l}{\mathrm{x}}_{\mathrm{i}}\in \mathrm{A}\Rightarrow {\mathrm{x}}_{\mathrm{i}}\in \mathrm{E},{\mathrm{x}}_{\mathrm{i}}\in \mathrm{F}\\ \text{ (or) }{\mathrm{x}}_{\mathrm{i}}\in \mathrm{E},{\mathrm{x}}_{\mathrm{i}}\notin \mathrm{F}\left(\text{ or }\right){\mathrm{x}}_{\mathrm{i}}\notin \mathrm{E},{\mathrm{x}}_{\mathrm{i}}\in \mathrm{F}\\ \text{ (or) }{\mathrm{x}}_{\mathrm{i}}\notin \mathrm{E},{\mathrm{x}}_{\mathrm{i}}\notin \mathrm{F}\\ \therefore \mathrm{n}\left(\mathrm{S}\right)=4^{10}\end{array}$
[each x₁ can be taken into sub-sets E, F in 4 ways]