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May 1st, 2018, 03:47 AM  #1 
Member Joined: Nov 2012 Posts: 74 Thanks: 1  Probability
The question is “a 5card poker hand is selected from a deck of 52 playing cards. How many different poker hands contain exactly one pair only?” My answer is (52C1 x 48C1 x 44C1 x 40C1 x 3C1) / 5! = 109 824 However, the suggesting answer is 10 times more, which is 1 098 240 Is there any mistakes for my calculation? 
May 1st, 2018, 11:37 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
Let the deck contain $v$ values of $s$ suits and we deal an $h$ card hand. Our values for this problem are $v=13,~s=4,~h=5$ I'm seeing that the formula for the number of hands with only a single pair is $n = \dfrac{s^{h2}\dbinom{s}{2} \prod \limits_{k=0}^{h2}(vk)}{(h2)!}$ for our current problem this does lead to $n=1098240$ You're neglecting for the variety of suits the pair can take. Last edited by romsek; May 1st, 2018 at 11:41 AM. 
May 1st, 2018, 12:29 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
Ok I have a good way to look at this now. Pick a pair. There are $v \dbinom{s}{2}$ different pairs. Now place them in the hand. There are $h\cdot (h1)$ ways of doing this as the two cards are distinguishable. Now pick the remaining $h2$ cards. There are $\prod \limits_{k=1}^{h2} s(vk)$ ways of doing this So all told the number of ordered hands with 1 pair is given by $v \dbinom{s}{2} h(h1) \prod \limits_{k=1}^{h2} s(vk) = \dbinom{s}{2}h(h1)s^{h2}\prod \limits_{k=0}^{h2} (vk)$ Now as we don't care about the order of the hand at all we divide all this by $h!$ to obtain $n1p = \dfrac{\dbinom{s}{2}h(h1)s^{h2}\prod \limits_{k=0}^{h2} s(vk)}{h!} = \dfrac{\dbinom{s}{2}s^{h2}\prod \limits_{k=0}^{h2} s(vk)}{(h2)!}$ Last edited by romsek; May 1st, 2018 at 12:32 PM. 
May 1st, 2018, 04:48 PM  #4 
Newbie Joined: Sep 2013 From: Reno,NV Posts: 14 Thanks: 1 Math Focus: Number Theory 
$\dbinom{13}{1} \dbinom{4}{2} \dbinom{12}{3} \dbinom{4}{1}^3$ This is from 'The theory of Gambling and Statistical Logic" by Richard A. Epstein The explanation is: The rank can occur in $\dbinom{13}{1}$ ways The cards forming the pair can be selected in $\dbinom{4}{2}$ ways The three other ranks can occur in $\dbinom{12}{3}$ ways A card of each rank can be selected in $\dbinom{4}{1}$ ways 
May 1st, 2018, 10:48 PM  #5 
Member Joined: Nov 2012 Posts: 74 Thanks: 1 
I think I know what’s wrong with my calculation. It seems that I cannot assume different suits of the same rank could be eliminated from the selection after each pick. That’s why it is incorrect to write 52C1 x 48C1.... What about this question: there are five classes. A committee consists of 15 members, with 3 representatives from each class. In how many ways can 3 members be selected from the committee if the 3 selected members are from different classes? There are two ways to approach the answer for me Approach 1: 5C3 x (3C1)^3 = 270 Approach 2: (15C1 x 12C1 x 9C1)/3! = 270 Although both approaches give the same result, is it the second approach not a correct one to do so? 

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