course Mth 152 ޔx־y͍H~assignment #006
......!!!!!!!!...................................
16:21:27 `q001. Note that there are 8 questions in this assignment. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's?
......!!!!!!!!...................................
RESPONSE --> There are four 5's in a deck of cards. C(4,2)=6 There would be 48 cards that are not 5's. We want 3 of those cards. C(48,3) = 17296. 6 * 17296 = 103,776 ways to get a hand containing exactly two 5's. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:21:52 In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C(4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's.
......!!!!!!!!...................................
RESPONSE --> Correct answer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:26:34 `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's?
......!!!!!!!!...................................
RESPONSE --> C(4,2) = 6 C(4,2) = 6 There are 44 cards left to choose from. We only need one of those cards. C(44,1) = 44 6 * 6 * 44 = 1,584 ways to get a hand containing two 5's and two 9's. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:26:57 There are C( 4, 2) ways to select exactly two 5's and C(4, 2) ways to select exactly two 9's. There are 44 remaining cards which are neither 9 nor 5. The total number of possible ways is therefore C(4, 2) * C(4, 2) * 44.
......!!!!!!!!...................................
RESPONSE --> Correct answer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:30:21 `q003. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three 9's?
......!!!!!!!!...................................
RESPONSE --> C(4,2) = 6 for the 5's C(4,3) = 4 for the 9's C(44,0) = 1 since you would have no remaining choices. 6 * 4 * 1 = 24 ways to get a full house consisting of two 5's and three 9's confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:30:41 There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3).
......!!!!!!!!...................................
RESPONSE --> Correct answer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:33:22 `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards?
......!!!!!!!!...................................
RESPONSE --> C(4,2) = 6 for two 5's C(12,3) = 220 for the face cards 6 * 220 = 1,320 ways to get a full house consisting of two 5's and three identical face cards confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:36:11 There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3).
......!!!!!!!!...................................
RESPONSE --> I was thinking that I would have to account for the fact that there are 12 face cards in the deck. We can only use 4 because there are four of each card, right? self critique assessment: 2
.................................................
......!!!!!!!!...................................
16:40:18 `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another?
......!!!!!!!!...................................
RESPONSE --> C(4,2) = 6 C(4,3) = 4 6 * 4 = 24 ways to get a full house consisting of two of one denomination and three of another denomination. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:41:07 For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses. There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind. Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses.
......!!!!!!!!...................................
RESPONSE --> I didn't finish the problem out, but I see how to arrive at the correct solution. self critique assessment: 2
.................................................
......!!!!!!!!...................................
16:53:30 `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit?
......!!!!!!!!...................................
RESPONSE --> There are 13 cards of the same suit. C(13,5) = 1,287 I'm not sure where to go from here. confidence assessment: 0
.................................................
......!!!!!!!!...................................
16:54:16 There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes.
......!!!!!!!!...................................
RESPONSE --> All I had to do was multiply my answer by 4. That makes sense! self critique assessment: 2
.................................................
......!!!!!!!!...................................
16:58:17 `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9?
......!!!!!!!!...................................
RESPONSE --> There are four possible cards for each number. 4 * 5 = 20 ways to get a straight consisting of one each of the denominations 5, 6, 7, 8, and 9. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:58:36 There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9.
......!!!!!!!!...................................
RESPONSE --> Correct answer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:02:59 `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'?
......!!!!!!!!...................................
RESPONSE --> If the ace can be high or low, there are 10 choices for the first card of the straight. The first card would have to be ace through 9. Since there are four suits, 10 * 4 = 40. For the remaining cards, there would only be four choices. 40 * 4 * 4 * 4 * 4 = 10,240 ways to get a straight of consecutive denominations with the ace being either high or low. confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:03:53 There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights.
......!!!!!!!!...................................
RESPONSE --> Correct answer. self critique assessment: 3
.................................................