Given a 50 card deck with cards numbered from 1 through 10 in each of 5 suits, how many 5 card hands are there that include exactly one pair of two cards that have the same numeric value?
The double can be any number from 1 to 10. 10 choices
for each of those there is 5C2 = 10 ways to chose the 2 that will be doubles.
now one the double suit is chosen there is 45 ways to chose the 3rd card
40 ways to chose the 4th card
35 ways to chose the last card
so I have
10*10*45*40*35 = 6 300 000 ways.
Sorry, Melody. I know this is a slightly hard problem, at least for me. I looked over your working to comprehend it, and it looks a bit like what I had done. My answer had only a slight arithmetic error, and hence the difference.
However, both of our answers appear to be wrong! Here is the answer I recieved from my source:
There are 10 options for the numeric value of the two cards in the pair. Once we selected it, there are C(5,2)=10 ways to pick which two suits the cards in the pair will come from. There are then C(9,3) = 84 ways to choose the numbers on the final three cards, and 5 choices for the suit for each card, giving us a total of 10*10*84*5³ = 1,050,000 ways.
Thanks for showing me that.
Your souce is right.
I have double counted.
For the last 3 cards I have counted order where I should not have.
I mean I have counted 345 separately from 354 etc. where they are both the same.
I could have fixed this by dividing my answer by 3! = 6
6 300 000/6 = 1050000 And now the answers are the same