How many possible ways there are to choose 5 cards from a Standard 52-card deck, in a way that between those 5 cards there will be exactly three cards with the same number?
In a Standard 52-card deck there are 4 suits and 13 enumerable cards.
1º) Choose a number between 1 and 13. You can do that in ${13\choose 1}$ ways.
2º) Choose 3 from the 4 suits. You can do that in ${4\choose 3}$ ways
3º) Now get your three cards with the number that you've decided. You can do that in only one way.
4º) Finally, choose 2 cards from the remaining ($52-3=49$) cards. You can do that in ${49\choose 2}$ ways. Therefore, by the multiplication principle: $$ {13\choose 1}{4\choose 3}{49\choose 2} $$ Is it correct? I'm a complete beginner in combinatorics... Any help is highly appreciated.
Thanks!
This is not correct, since you want exactly 3 cards that have the same number. The correct answer should be $\binom{13}1\binom43\binom{48}2$ since you can not choose the same number to the other 3 cards. Even if the question is AT LEAST $3$ cards, then still there are double counting. For example, 1,1,1,1,13 is counted 4 times.