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please refer the image if my answer is correct

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tarun8572
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    A good way for you to settle the question is to work it for a smaller deck. Suppose there are only $4$ cards in the deck. What's the answer? What does that tell you? – lulu Jan 30 '20 at 12:04
  • @lulu is 52C3 is correct or 52C3*(1/6) – tarun8572 Jan 30 '20 at 12:07
  • I understand that you just want someone to hand you the answer, but try my example. That one is small enough to work by hand. – lulu Jan 30 '20 at 12:22
  • thanks, @lulu I think I figured it out There are 4C3 ways to choose which suit goes first and stuff so that's 4. – tarun8572 Jan 30 '20 at 12:27
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    Exactly!. In this case, $\binom 43$ isn't even divisible by $6$. In general $\binom nk$ refers to exactly what you want, the number of ways to choose an unordered subset from $n$ distinct objects. – lulu Jan 30 '20 at 12:30
  • are all distinguishable ? –  Jan 30 '20 at 12:32
  • thank u very much @lulu so NONE of the above is the answer – tarun8572 Jan 30 '20 at 12:34
  • No...look at your example and read my comments. For $n=4, k=3$ you got $\binom 43$. If you are still confused, work the same problem with $n=5$. And study the definition of the binomial coefficient. – lulu Jan 30 '20 at 12:35

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Choosing 3 from 52 is $\binom{52}{3}$. Given that $3!=6$, it sounds like you're thinking of dividing by $3!$ in order to ensure order doesn't matter. However, the $3!$ division is already included in $\binom{52}{3}$: $\binom{52}{3} = \frac{52!}{3!49!} = \frac{\frac{52!}{49!}}{3!}$. The numerator $\frac{52!}{49!}$ is actually $P(52,3)$, choosing $3$ from $52$ when order matters.

It seems like what you intended was to choose $\frac{P(52,3)}{3!}$. We have $\frac{P(52,3)}{3!} = \binom{52}{3}$, not $\frac{P(52,3)}{3!} = \frac{\binom{52}{3}}{3!}$

BCLC
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    is 52C3 is the right answer for different cards – tarun8572 Jan 30 '20 at 12:51
  • @tarun8572 Yes. Actually in my answer, I kind of assume the cards are different. (If the cards weren't different, i.e., if we allow repetition, then I think... the answer is $(52)^3$ if order matters and then $(52)^3$ divided by something if order doesn't matter.) – BCLC Jan 30 '20 at 12:53