STATS 300 - Section 5.5 Homework
5.5.35 - A salesperson must travel to ten cities to promote a new marketing campaign. How many different trips are possible if any route between cities is possible?
10!
5.5.45 - Suppose 51 cars start at a car race. In how many ways can the top 3 cars finish the race? 124950
nPr=n!/(n−r)! The value of n is the number of cars to choose from, so n = 51. The value of r is the number of cars to choose, so r = 3. 51P3 = 51!/(51-3)! = 124950
5.5.7 - Find the value of the factorial. 13! = 6,227,020,800
(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800
5.5.55 - How many different 10-letter words (real or imaginary) can be formed from the following letters? 907200 X, D, A, Z, A, K, D, W, E, Q
10!/(1!)(2!)(2!)(1!)(1!)(1!)(1!)(1!) = 907200 (10 total, 2 d's and 2 a's)
5.5.57 - A golf-course architect has four linden trees, four white birch trees, and three bald cypress trees to plant in a row along a fairway. In how many ways can the landscaper plant the trees in a row, assuming that the trees are evenly spaced? 11550
Determine the total number of trees => n = 11 Let n1 be the number of linden trees, n2 be the number of white birch trees, and n3 be the number of bald cypress trees. The number of linden trees is n1=4. The number of white birch trees is n2=4. The number of bald cypress trees is n3=3. n!/(n1!)(n2!)(n3!) = 11!/(4!)(4!)(3!) = 11550
5.5.19 - Find the value of the combination. 7C5 = 21
combination formula - nCr=n!/r!(n−r)! 7C5 = 7!/5!(7-5)! = 21
5.5.63 - Suppose a shipment of 170 electronic components contains 4 defective components. To determine whether the shipment should be accepted, a quality-control engineer randomly selects 4 of the components and tests them. If 1 or more of the components is defective, the shipment is rejected. What is the probability that the shipment is rejected? 0.0867
nCr=n!/r!(n−r)! Since there are 170−4=166 non-defective components, the number of ways of selecting 4 non-defective components is 166C4. 166C4 = 166!/4!(166-4)! = 30507895 The number of ways of selecting 4 components from 170 components is 170C4. Calculate 170C4. 170C4=170!4!(170−4)! = 33585370 P(0 defective) = 166C4/170C4 = 30507895/33585370= 0.9084 P(1 or more defective) = 1 − P(0 defective) = 1 - 0.9084 = 0.0916
5.5.51 - How many different simple random samples of size 5 can be obtained from a population whose size is 48?
nCr=n!/r!(n−r)! The value of n is the number of people to choose from, so n = 48. The value of r is the number of people to choose, r = 5 48C5 = 48!/5!(48-5)! = 1712304
5.5.44 - A license plate is to consist of 4 digits followed by 2 uppercase letters. Determine the number of different license plates possible if the first and second digits must be odd, and repetition is not permitted. 728000
The first digit in the license plate must be odd. There are 5 odd digits from 0-9 so there are 5 possible choices for the first digit. The second digit on the license plate must also be odd, and repetition is not allowed. Since there are 5 odd digits, and one digit is used for the first choice, there are 4 possible choices for the second digit. The third digit on the license plate can be either even or odd, but repetition is still not allowed. Since repetition is not allowed and 2 digits have already been used, that leaves 8 possible choices for the third digit. The number of possible choices for the fourth digit is found in the same manner. The possible number of choices for each digit is as shown. 5/D 4/D 8/D 7/D Then consider the 2 letters in the code. Since there are 26 letters in the alphabet, there are 26 possible choices for the first letter. Repetition is not allowed, so there are 25 choices for the second letter. The possible number of choices for each digit as well as each letter is as shown. 5/D 4/D 8/D 7/D 26/L 25/L Use the Multiplication Rule of Counting to determine the number of possible license plates. (5)(4)(8)(7)(26)(25) = 728000
5.5.43 - A certain three-cylinder combination lock has 55 numbers on it. To open it, you turn to a number on the first cylinder, then to a second number on the second cylinder, and then to a third number on the third cylinder and so on until a three-number lock combination has been effected. Repetitions are allowed, and any of the 55 numbers can be used at each step to form the combination. a. How many different lock combinations are there? 166375 b. What is the probability of guessing a lock combination on the first try? 1/166375
a. Formula = n^r Thus, the number of possible three-number lock combinations is given by the following calculation. 55•55•55=55^3 = 166375 b. Therefore, the probability that the correct code is guessed on the first try is 1/55^3 = 1/166375
5.5.29 - Complete sections a and b a. List all the combinations of five objects m, l, n, k, and p taken two at a time. b. What is 5C2? 10
a. ml, mn, mk, mp, ln, lk, lp, nk, np, kp b. 5C2 = 5!/2!(5-2)! = 10
5.5.69 - Suppose you have just received a shipment of 19 modems. Although you don't know this, 3 of the modems are defective. To determine whether you will accept the shipment, you randomly select 6 modems and test them. If all 6 modems work, you accept the shipment. Otherwise, the shipment is rejected. What is the probability of accepting the shipment?
nCr=n!/r!(n−r)! Since there are 19 − 3 = 16 non-defective modems, the number of ways of selecting 3 non-defective modems is 16C6 = 16!/6!(16-6)! = 8008 Calculate 19C6. 19C6 = 19!/6!(19−6)! = 27132 P(0 defective) = 16C6/19C6 = 0.2951
5.5.49 - A lottery exists where balls numbered 1 to 16 are placed in an urn. To win, you must match the five balls chosen in the correct order. How many possible outcomes are there for this game?
nPr=n!/(n−r)! The value of n is the total number of distinct objects to choose from, so n = 16. The value of r is the number of objects to be chosen, so r = 5. 16P5 = 16!/(16-5)! = 524160
5.5.11 - Find the value of the permutation. 3P1 = 3
permutation formula - nPr=n!/(n−r)! 3P1 = 3!/(3-1)! = (1)(2)(3)/(3-1)! =6/(1)(2) = 3
