"How many ways are there to choose four cards of different suits and different values from a deck of 52 cards ? "
I solved this question in this way :- "So, for the first card we have 52 options to choose from .for eg. king of diamonds . so for the next card we can't choose from the suit of diamonds with the value of a king . Hence, there are 36 options to choose the second card . [ 52-13(suit of diamond)=39-3(rest of the kings)]. for eg. queen of hearts has been chosen as second. So, for choosing third card we have 21 options to choose from . [ 36-13(suit of hearts)=23-2(rest of the queens)] . for eg. ace of spades has been chosen as third . Now, for the last card we have 7 options to choose from . [ 21-13(suit of spades)=8-1(rest of the ace)] . Hence, the total no. of ways should be :- 52x36x21x7 = 275184 ways " I was told that this logic is wrong and there will be some overcounting . But, I couldn't understand why . So, can someone please point out the flaw in this method .
If we want to do it your way, two figures need correction, and you need to divide by $4!$ because the cards could have been chosen in various orders,
thus $\;(52\times 36\times22\times10)\div 4!$
As you can see, the approach you adopted is complex and error prone, and a much simpler way is to simply choose values of cards suit by suit,
thus $\;13\times12\times11\times10$