Four cards are randomly drawn from a pack of 52 cards. Find

My solution:

Bunuel, thank you for your answer.

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. T

Wanted just to correct myself. The above argument is wrong since: 52 x 48 x 44 x 40 is a case of variations without repetition. Therefore, the set won't contain cards of the same denomination.

I think you are right:

\(\frac{4^4*C^4_{13}}{C^4_{52}}\) would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -\(C^4_{13}\);
Multiplying by \(4^4\) as each card from the above string can be of four suits (\(4*4*4*4=4^4\));
Total # of ways to choose 4 cards out of 52 - \(C^4_{52}\).

Or another way:

We can choose any card for the first one - \(\frac{52}{52}\);
Next card can be any card but 3 of the denomination we'v already chosen - \(\frac{48}{51}\) (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left);
Next card can be any card but 3*2=6 of the denominations we'v already chosen - \(\frac{44}{50}\) (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left);
Last card can be any card but 3*3=9 of the denominations we'v already chosen - \(\frac{40}{49}\);

\(P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49}\) - the same answer as above.

Hope it helps.

No posting of PS/DS questions is allowed in the main Math forum.[/warning]

Why wouldn't the probability be 1 * 13/51 * 13/50 * 13/49 ? We know there are 4 suits, and the probability of getting any card in one suit is 13/# of cards in the deck.

In your solution, you found the # of ways to choose 4 different denominations out of 13, I believe that is incorrect. Each suit(denomination) has 13 cards, we're looking for the # of ways to draw 1 card of a different suit each time, up to 4 draws. Correct me if I'm wrong.

I think you are right:

\(\frac{4^4*C^4_{13}}{C^4_{52}}\) would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -\(C^4_{13}\);
Multiplying by \(4^4\) as each card from the above string can be of four suits (\(4*4*4*4=4^4\));
Total # of ways to choose 4 cards out of 52 - \(C^4_{52}\).

Or another way:

We can choose any card for the first one - \(\frac{52}{52}\);
Next card can be any card but 3 of the denomination we'v already chosen - \(\frac{48}{51}\) (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left);
Next card can be any card but 3*2=6 of the denominations we'v already chosen - \(\frac{44}{50}\) (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left);
Last card can be any card but 3*3=9 of the denominations we'v already chosen - \(\frac{40}{49}\);

\(P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49}\) - the same answer as above.

How did \(\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}\) become equal to \(\frac{4^4*13*12*11*10}{52*51*50*49}\) here?

Take 4 common from the numerator and the result would be \(4*\frac{13*12*11*10}{52*51*50*49}\).

I think we we should not multiply it with 4 since we are not finding number of ways just probability so order doesnot matter..

13/52 * 13/51*13/49*13/48 should be the answer ...

Please correct me if i am wrong.