1. [-/1 Points] DETAILS Evaluate the number. A2. 21 Need Help? 2. [-/1 Points] DETAILS Evaluate the number. A6. 2] Need Help? 3. [-/1 Points] DETAILS Evaluate the number. CT9, 6) Need Help? 4. [-/1 Points] DETAILS Evaluate the number. CTE, S) Need Help? 5. [-/1 Points] DETAILS How many ordered lists are there of eight items chosen from fourteen? | ordered lists6. [-/1 Points] DETAILS How many ordered sequences are possible that contain six objects chosen from fifteen? ordered sequences Need Help? 7. [-/1 Points] DETAILS How many unordered sets are there of four Items chosen from eight? set Need Help? 8. [-/1 Points] DETAILS How many four-letter sequences are possible that use the letters a, o, j, w once each? sequences Need Help?9. [1/9 Points] DETAILS PREVIOUS ANSWERS This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise How many four-letter sequences are possible that use the letters d, l, n, to v at most once each? Step 1 We wish to determine the number of four-letter sequences that are possible if the letters d, I, n, t, and v are the only choices and if each letter is used at most once each. Here a four-letter sequence such as d-I-n-t would be one such sequence, while different orderings of the same four letters such as d-n-t-l, d-t-lon, etc. would be counted as entirely different four-letter sequences. Therefore, the order in which the letters are picked |matters matters Step 2 Since the order that the letters are picked matters, the number of possible sequences can be counted using a permutation. Recall that a permutation of n items taken & at a time is an ordered list of r items chosen from n, and is written as P(n, r). The possible letters are d, I, n, t, and v, and we wish to form a four-letter sequence. Therefore, we must calculate a permutation of items taken at a time. In other words,10. [-/1 Points] DETAILS Your international diplomacy trip requires stops in Thailand, Singapore, Hong Kong, and Bali. How many possible itineraries are there in which the last stop is Thailand? HINT [See Examples 1 and 2.] itineraries Need Help? Read It 11. [-/1 Points] DETAILS Your international diplomacy trip requires stops in Thailand, Singapore, Hong Kong, Laos, and Bali. How many possible itineraries are there? HINT [ See Examples 1 and 2.] itineraries Need Help? Read It12. [-/11 Points] DETAILS MY NO This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Poker Hands A poker hand consists of 5 cards from a standard deck of 52. Find the number of different poker hands of the specified type. two pairs (two of one denomination, two of another denomination, and one of a third) For those unfamiliar with playing cards, here is a short description. A standard deck consists of 52 playing cards. Each card is in one of 13 denominations: ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, jack (]), queen (Q), and king (K), and in one of four suits: hearts (v), diamonds (+), clubs (@), and spades (). Thus, for instance, the jack of spades, Je, refers to the denomination of jack in the suit of spades. The entire deck of cards is as shown below. AY 2V 3V 6Y SV gv 10Y JY OV KY A+ 24 6+ 84 104 J+ 0+ K+ A& 24 14 84 10% J+ 0+ Ke 36 44 64 74 84 10% Je Q. Step 1 Recall that the entire deck of cards is as follows. AY 6Y 10Y JY QV KY A+ 24 34 8+ 94 10+ J+ Q+ K. At 24 54 84 94 104 J+ 0+ Ks 2+ 10% Je Q. KeStep 1 Recall that the entire deck of cards is as follows. SV SY gV 10Y JY QV KY At 7+ 8+ 94 10+ J+ Q+ K+ 24 34 54 84 94 104 Je 0+ K- ge 10% Je Q. The goal is to determine the number of different five-card poker hands that consists of two pairs. In other words, we need to find the number of ways to pick 2 cards of one denomination, 2 cards of another denomination, and a final card that belongs to neither of the first two denominations. Our decision algorithm is a sequence of four steps. Step 1: Select 2 denominations for the pairs. Step 2: Select 2 cards from one selected denomination, Step 3: Select 2 cards from the other selected denomination. Step 4: Select 1 card that belongs to neither of the previously selected denominations. We begin with Step 1. There are 13 possible denominations and we will pick 2 of them, Because the order that we make this selection does not matter, we will use combinations to find the number of ways to pick 2 denominations out of 13. Therefore, in each case we will be using combinations to count the number of ways to take n items taken r at a time, which is calculated using the following formula. C(or) = r!(n - r)! Here we have n = 13 and / = 2. Substitute these values into the formula and simplify. n! C(n, r) =. r!(n - )! C(13, 2) = 2! (13 - 2)! In other words, there are possible ways to pick 2 of the 13 denominations for the pairs. Submit |Skip (you cannot come back)