$\begin{array}{1 1}(A)\;\large\frac{13C_3\times 13C_1}{32C_4}\\(B)\;\large\frac{13C_2\times 13C_1}{32C_3}\\(C)\;\large\frac{10C_3\times 10C_1}{12C_4}\\(D)\;\large\frac{9C_3\times 10C_1}{32C_4}\end{array} $

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- $nC_r=\large\frac{n!}{r!(n-r)!}$

Step 1:

Given deck of 52 cards

Cards have 13 diamond cards and 13 spades

$\therefore$ Totally 4 cards are to be obtained

$\Rightarrow$ Number of ways selecting 4 cards out of 52 cards

n(S)=$52C_4$

Step 2:

Let E be the event obtaining 3 diamonds and 1 spade

$\Rightarrow n(E)=13C_3\times 13C_1$

$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$

$\Rightarrow \large\frac{13C_3\times 13C_1}{52C_4}$

Hence (A) is the correct answer.

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