probability

Question

From a pack of $52$ cards, all the face cards are removed and four cards are drawn. Then the probability that they are of different suit and different denomination is

Answer 1

There are 12 face cards.

Remaining cards = 52 -12 = 40 cards

we can select 4 cards out of 40 cards in $_{4}^{40}\textrm{C}$ ways

Now, each suit will have 10 cards after removing 3 face cards of each suit

1. number of ways of selecting a card from particular suit is $_{1}^{10}\textrm{C} = 10 ways$

2. No of ways selecting the second card from another suit and of the different denomination(one card is fixed)  $_{1}^{9}\textrm{C} = 9 ways$

3. No of ways selecting the third card from another suit and of the different denomination(two cards are fixed)  $_{1}^{8}\textrm{C} = 8ways$

4. No of ways selecting the fourth card from another suit and of the different denomination(three cards are fixed)  $_{1}^{7}\textrm{C} = 7ways$

So, the total number of ways of selecting four cards of different suit and denomination is $10*9*8*7 =  5040 ways$

$Probability = 5040 / (_{4}^{40}\textrm{C}) $

                   $= 5040/91390$

                   $=504/9139$

                   $= 0.0551$

Comments

Thank u for the answer.

But I have a doubt. Why r u selecting the first card in 10 ways. first card can be any one of the forty cards.

so it can be selected in 40 ways.

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if am selecting from 1 particular suit

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@pankaj, since we have 4 suit. why dont you consider 4 different suit choice like first card can be from any suit

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pankaj You are selecting the 1^st  card from a particular suit because order does not matter here. Am I right. ?