What is the probability that a random bridge hand of 13 cards contains cards of all four suits

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The first card does not matter. The second card has to be one of the 39 of a different suit of the 51 still left. Then, one of the 26 of the 2 other suits of the 50 still left. finally, one of the 13 the 49 that at left.

Then the pattern is that with each card, the bottom number goes down by one and the top number is 12 for four time, 11 for four times, 10 for four times etc.

So the bottom number is 51!, that is "51 factoral."

The top number is 39 X 26 X 13 X (12 to the fourth) times (11 to the fourth) etc.

I do not have a calculator. Figure it out.