course mth 152 ¤gÿóŽÝùÄËÃÔúàyÞœKÎqéªÙÒŠ¦†assignment #006
......!!!!!!!!...................................
09:56:21 `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 52 cads in a deck you want 2 fives C(52,2) 52!/2!*50!= 52*51/2*1=1326 confidence assessment: 1
.................................................
......!!!!!!!!...................................
10:11:15 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 --> I understand that there are 52 cards in a deck 48 are not fives because there are 4 fives In our hand we have 2 fives out of 4 fives so C(4,2) there are 48 non fives and we need the three that are not five to make our hand so C(48,3) then multiply because order doesnot matter C(4,2)*C(48,3) self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:26:05 `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 --> there are 52 cards we need 2 fives and 2 nines there are four fives and four nines all together that leaves 44 cards that are not five or nine C(4,2) is the number of fives C(4,2) is nines C(4,2)*C(4,2) is 2 fives and 2 nines that left 44 other cards C(44,1) we have 44 cards but need 1 to make our hand C(4,2)*C(4,2)*C(44,1) is 2 fives and 2 nines confidence assessment: 1
.................................................
......!!!!!!!!...................................
10:27:49 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:31:53 `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 --> there are 52 cards we need 2 fives so there are four fives C(4,2) is the fives we need 3 nines and there are four so C(4,3) that leaves 47 that we wont need so C(47,0) C(4,2)*C(4,3)*C(47,0) confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:34:07 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 --> I shouldnt have multiplied the remaining cards because we already had the five cards to make a hand self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:42:36 `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 --> there are four fives so we need 2 C(4,2) there are kings, queens and jacks so there are 12 possible face cards we need 3 identical C(12,3) C(12,3)*C(4,2) confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:47:24 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 --> so I dont need to use the 12 possible face cards I just do the number of different faces so it would be 3*C(4,2)*C(4,3) self critique assessment: 2
.................................................
......!!!!!!!!...................................
11:11:16 `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 --> If we pick 2 seven and 3 two then there are C(4,2) and C(4,3) C(4,2)*C(4,3) to get this full house you have four different suits and 13 different values of cards you need two different suitsandf since this is an arrangement you use P so P(13,2) Multiply all C(4,2)*C(4,3)* P(13,2) confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:14:41 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:30:18 `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 four suits and a flush is of the same suit all hearts for instance a royal is different so you have 12 cards in each suit and you need five of the same suit confidence assessment: 0
.................................................
......!!!!!!!!...................................
11:33:06 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 --> I was counting 12 because I thought that ace was not counted self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:40:39 `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 fives four sixes four sevens and four eights and four nines if suit doesnt matter c(4,1) is 5s C(4,1) is 6s C(4,1) is 7s C(4,1) is 8s C(4,1) is 9s C(4,1)*C(4,1) * C(4*1)*C(4,1)*C(4,1) confidence assessment: 1
.................................................
......!!!!!!!!...................................
11:42:42 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:51:36 `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 --> there are 13 cards of each suit if the ace is high 2-A 13 ways if the ace is low A-K 14 ways if a is high and low need 5 cards of any suit 4*C(14,5) confidence assessment: 1
.................................................
......!!!!!!!!...................................
11:59:09 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 --> I was thinking that ace was high then it was low I understand that there are 10 cards if you dont count j k q so there are four of each number so C(4,1) =4 and there are 5 cards so 4^5 for one hand and there could be 10 possible hands so 10*4^5 self critique assessment: 2
.................................................