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Four cards are drawn from a pack of 52 cards. What is the chance that no two cards are of equal value? Each belongs to a different suit?
My try--
Total probability =$^{52}C_4$
In the pack of 52 cards there are 4 aces, 4 1s and so on. So, there will be 13 groups of 4 cards containing same value.
$\therefore$ Probability that no two cards are of equal value =$\frac{^{13}C_4\times ^4C_1}{^{52}C_4}=\frac{44}{4165}$
But given answer is=$\frac{2816}{4165}$
I think I did it wrong, please help me. I also couldn't solve the 2nd problem.

Ankita Pal
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    The probability you calculated of $\binom{13}{4}\times\binom{4}{1}/\binom{52}{4}$ is the probability that every card is of different rank but every card is of the same suit. If you wanted all cards to be different rank and different suit, that should have been a $4!$, not a $\binom{4}{1}$. If you didn't care about suits being repeated then that should have been a $4^4$... or if you absolutely insist... $\binom{4}{1}\times\binom{4}{1}\times\binom{4}{1}\times\binom{4}{1}$ (but this just makes it harder to read and is annoying) – JMoravitz Oct 08 '20 at 15:56
  • As for the question of asking the probability that all four cards are of different suit but you don't care about repeated ranks, that is covered by what I linked this as a duplicate of. – JMoravitz Oct 08 '20 at 15:59
  • @JMoravitz are you saying it will be $\frac{^{13}C_4\times 4!}{^{52}C_4}$? But, it doesn't match the answer. – Ankita Pal Oct 08 '20 at 16:12
  • All different ranks and all different suits: $\binom{13}{4}\cdot 4!/\binom{52}{4}$. All different ranks, all the same suit: $\binom{13}{4}\cdot 4/\binom{52}{4}$. All different ranks, no restriction on suits: $\binom{13}{4}4^4/\binom{52}{4}$. Read what I wrote more carefully. Continuing: All the same rank: $13/\binom{52}{4}$. No restriction on ranks all different suits: $13^4/\binom{52}{4}$, and so on... There are many similarly sounding questions with slightly altered conditions. Do not confuse one question for another. – JMoravitz Oct 08 '20 at 16:28

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