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1. In poker, a hand with 5 cards has two pairs if it contains two cards of one rank, two cards of a different rank, and a fifth card that does not match either of the first two ranks. (For example, the hand JJ774 has two pairs.) The order of the cards does not matter (so JJ774 and 7J74J are the same). How many hands with 5 cards have two pairs? (All the cards are taken from a standard deck of 52 cards.)

 

2. A committee with nine people gather to have a meeting. The members sit around a circular table. Two of the committee members, Rick and Morty, want to sit so that there is exactly one person between them. How many different seatings are possible? (Two seatings are considered the same if one can be rotated to obtain the other.)

 

I don't know how to solve either of these, and I don't really understand the first one. 

 Dec 23, 2019
 #1
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1. First we decide what ranks we will use. Then we will pick suits for all of the cards.

We choose the two paired ranks in \({13\choose 2}=78\) ways and the remaining rank in \({11\choose 1}=11\) ways. Then we choose the suits for these cards in \({4\choose 2}{4\choose 2}{4\choose 1}=144\) ways. This gives a total of \(78\cdot 11\cdot 144 = \boxed{123552}\) two pair hands.

 Dec 23, 2019
 #2
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2. We seat Rick first. He has 9 choices. We seat Morty next. He has 2 choices: two spots to the left of Rick or two spots to the right of Rick. Then, the remaining seven members can sit in 7! ways. This appears to give a total of 9*2*7! seatings. However, because two seatings are considered the same if one is a rotation of the other, each distinct seating is counted 9 times in this product, once for each possible location of Rick. So, we must divide by 9 to give a total of \(2\cdot 7! = 2\cdot 5040 = \boxed{10080}\) seatings.

We might have instead started by noting that it doesn't matter where Rick sits. He starts by sitting anywhere, and we count the number of ways the members sit relative to Rick. Morty still has 2 choices: two spots to the left of Rick or two to the right of Rick. The remaining seven members then can be seated in 7! ways, for a total of \(2\cdot 7! = 2\cdot 5040 = \boxed{10080}\) seatings, as before.

 Dec 23, 2019
 #3
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Thank you so much!

Guest Dec 23, 2019
 #4
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no prob :)

Guest Dec 23, 2019

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