Suppose 9 cards are numbered with the 9 digits from 1 to 9

[r-soy]

I try to solve this queation and see >>

suppose 9 cards are numbered with the 9 digits from 1 to 9 . A 3 card hand is dealt , 11 card at time . How many hands are possible where :

I ) order is taken into considreation
II ) order is not taken into considreation

I ) 9P3 = 504
II) 9C1 = 9

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Q2 : From a standerd 52 card deack how many 5 card hands will have all face cards ? ( consider only Jaks , queens and kings to be the face card ?

13C3.13C2
= 22.308

 

[r-soy]

I want help were are you ?

 

[hgfalling]

For Q1, Part I, you are right.

For Part II, your answer is 9? But look, I can name ten different hands:

123
124
125
126
127
128
129
134
135
136

Hint: The only difference between part I and part II is that order doesn't matter.


For Q2, I'm not really sure what's going on. I see you're taking combinations out of 13, which is the number of ranks. But that's not right, because first of all, a deck has 52 cards in it and the cards are dealt without replacement. But you're not concerned with the 52 cards, only with the ranks that are face cards. There are 12 of those cards total. So the question is asking, "How many ways can you choose five cards out of 12 cards?"