Counting Principles [AutosavedCounting Principles [Autosaved].pptx].pptx
- 2. First counting principle • The fundamental counting principle or basic principle of counting is a method or a rule used to calculate the total number of outcomes when two or more events are occurring together. • This principle states that the total number of outcomes of two or more independent events is the product of the number of outcomes of each individual event. • The Fundamental Counting Principle states that if an event or decision has a possible outcomes or choices, and another event has b possible outcomes or choices, then the total number of possible unique combinations of outcomes between the two is a b ⋅ .
- 3. Basic Counting Principles: Sum Rule • When two events or decisions are independent of each other and cannot be done at the same time, the total number of choices are simply added to find the total possible choices between the two events or decisions. • In other words, the choices are mutually exclusive.
- 4. Count the number of possibilities when a coin is tossed 3 times. Solution: A coin toss can have two outcomes, either Heads(H) or Tails(T), and in case of tossing three coins simultaneously the total number of ways in which this can happen is, = Total Outcome of First Coin Toss × Total Outcome of Second Coin Toss × Total Outcome of Third Coin Toss = 2 × 2 × 2 = 8
- 5. Find the number of four-letter words with or without meaning, which can be made out of letters of the word ROSE, where the repetition of letters is not allowed. Solution: Number of words that can be formed from these four-letter words is equal number ways in which we can fill __ __ __ __ with letters R, O, S, E. Note that repetition is not allowed. The first place can be filled with any of the four letters, after that second place can only be filled by three letters because we have already used one and repetition is not allowed. Third place can only be filled by two letters and last place will be filled with the last remaining letter. So, number of ways in which we can do this are. 4 × 3 × 2 × 1 = 24. Note: If the repetition of the letters was allowed we could have always used four letters to fill each place. So 4 × 4 × 4 × 4 = 256.
- 6. Given 6 flags of different colors, how many different signals can be generated, if a signal requires the use of 2 flags one below the other? • Solution: • Here in each position we can use the different colors of flag we are given. So, in the first position we have 6 different choices to make to fill in the position of flag 1. So, in the second position we will have 5 positions to fill because we have already used one color. • So, total number of ways to fill = 6 × 5 = 30.
- 7. How many 2-digit even numbers can be formed from the digits 1, 2, 3, 4, and 5 if the digits can be repeated? Solution: There are five possibilities for putting numbers in each place since the numbers can be repeated. But a constraint is given in the questions which says that the number should be even. So, all the even numbers have an even digit as the last digit. In the given numbers, only 2 and 4 are two even numbers. So, at the unit’s place in the number, there are only two possibilities while their 5 possibilities for the tens place. So, Total number of possible even numbers = 5 × 2 = 10
- 8. How many positive divisors do 1000 = 23 53 have? Solution: The positive divisor of 1000 will be in form 2a 5b . Where, a and b will satisfy 0 ≤ a ≤ 3 and 0 ≤ b ≤ 3 It is clear that there are 4 possibilities of a and 4 possibilities of b. Hence, there are 4 × 4 = 16 Positive Integers of 1000.
- 9. suppose a bakery has a selection of 20 different cupcakes, 10 different donuts, and 15 different muffins. If you are to select a tasty treat, how many different choices of sweets can you choose from? • Because we have to choose from either a cupcake or donut or muffin (notice the “OR”), we have 20 + 10 + 15 = 45 treats to choose from. • suppose a bakery has a selection of 20 different cupcakes, 10 different donuts, and 15 different muffins — how many different orders are there? • Because we can choose treats from a selection of cupcakes and donuts and muffins (notice the “AND”), we 20 x 10 x 15 = 3,000 ordering options.
- 10. A computer password is to consist of two lower case letters followed by four digits. Determine how many passwords are possible if….. • A) Repetition of letters and digits is permitted • First create a spot for each object that needs to be placed. ____ ____ ____ ____ ____ ____ = L L # # # # • Next place numbers in each spot and multiply the numbers. • Each of the spot marked “L” will get a 26 above it as there are 26 letters. . • Each of the spot marked “#” will get a 10 above it as there are 10 digits (0,1,2,3,4,5,6,7,8,9) to choose from. 26 * 26 *10*10*10*10 = 6,760,000
- 11. • Repetition of letters and digits is NOT permitted First create a spot for each object that needs to be placed. ____ ____ ____ ____ ____ ____ = L L # # # # Next place numbers in each spot and multiply the numbers. • Each of the spot marked “L” will get a 26 above it as there are 26 letters. I need to increment down by one as repetition is NOT allowed. • Each of the spot marked “#” will get a 10 above it as there are 10 digits (0,1,2,3,4,5,6,7,8,9) to choose from. • I need to increment down by one as repetition is NOT allowed. 26 * 25 * 10 * 9 *8* 7 = 3,276,000
- 12. • The first letter must be a vowel (a, e, i, o, u) and repetition of letters is NOT permitted. The first digit cannot be a 0, and repetition of digits is NOT permitted. ____ ____ ____ ____ ____ ____ = different passwords V L # # # # • The first letter has 5 choices as it needs to be a vowel. • The second letter has 25 choices. • Repetition is not permitted. I can use any of the 25 letters not already used. • The first digit has 9 choices as it can be any of the 10 digits other than 0. {1,2,3,4,5,6,7,8,9} The second digit also has 9 choices as it can be the 0 and any of the 8 numbers not selected for the first number. The third and 4th digit will increment down 1 as repetition is NOT permitted. 5*25*9*9*8*7 = 567,000
- 13. 1. Bob, Sue, Larry, Sally and Fred are waiting in line to buy concert tickets. A) In how many different ways can they stand in line? 5* 4 * 3 * 2 * 1 = 120 How many ways can they line up if Larry insists on being in the middle? 4 * 3 * 1 * 2 * 1 = 24 If Sue insists on being first and Bob has to be last, how many ways? 1 * 3 * 2 * 1 * 1 = 6
- 14. • 2. The letters S, T, A, R, P are placed in a hat. A letter is removed one at a time from the hat and is not replaced. This process is repeated until all of the letters have been removed. How many different ways can this be done? • 5 * 4 * 3 * 2 * 1 = 120 • Solving using the permutation formula: n = 5 and r = 5 (“n” represents the number of letters in the hat, and “r” represents the total number of letters I am removing.) • nPr = 5P5 = 5! /(5−5)! = 5 4 3 2 1 /0! = 120 1 ∗ ∗ ∗ ∗ • Answer: 120 ways • The letters S, T, A, R, P are placed in a hat. A letter is removed one at a time from the hat and is not replaced. This process is repeated 3 times. That is 3 letters are removed and 2 are left in the hat. How many ways can this be done? • 5 * 4 * 3 = 60 • n = 5 and r = 3 (“n” represents the number of letters in the hat, and “r” represents the number of letters I am removing.) nPr = 5P3 = 5!/ (5−3)! = 5 4 3 2 1 /2! = ∗ ∗ ∗ ∗ 120 /2 Answer: 60 ways
- 15. • A club consisting of eight members wishes to randomly select a president, vice-president and secretary, how many different arrangements are possible? • Solved using the permutation formula as it is an ordered arrangement of 3 out of 8 objects: 8 3P 8! (8−3)! = 8! 5! = 8 7 6 5 4 3 2 1 5 4 3 2 1 = 8 7 6 = 336 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ • Solved using the Counting Principle (I solve this as a without repetition problem as I assume one person can’t hold two jobs.) ______ _______ _______ President VP Secretary • 8 * 7 * 6 = 336 Answer: 336 ways
- 16. • In how many ways 3 mathematics books, 4 history books, 3 chemistry books and 2 biology books can be arranged on a shelf so that all books of the same subjects are together. • Solution First we take books of a particular subject as one unit. Thus there are 4 units which can be arranged in 4! = 24 ways. • Now in each of arrangements, mathematics books can be arranged in 3! ways, history books in 4! ways, chemistry books in 3! ways and biology books in 2! ways. Thus the total number of ways = 4! × 3! × 4! × 3! × 2! = 41472.
- 17. • A student has to answer 10 questions, choosing atleast 4 from each of Parts A and B. If there are 6 questions in Part A and 7 in Part B, in how many ways can the student choose 10 questions? Solution The possibilities are: 4 from Part A and 6 from Part B or 5 from Part A and 5 from Part B or 6 from Part A and 4 from Part B. Therefore, the required number of ways is • 6C4 × 7C6 + 6C5 × 7C5 + 6C6 × 7C4 = 105 + 126 + 35 = 266.
- 18. • Three married couples are to be seated in a row having six seats in a cinema hall. If spouses are to be seated next to each other, in how many ways can they be seated? Find also the number of ways of their seating if all the ladies sit together. • Solution Let us denote married couples by S1 , S2 , S3 , where each couple is considered to be a single unit 1 st 2 nd 3 rd • Then the number of ways in which spouces can be seated next to each other is 3! = 6 ways. • Again each couple can be seated in 2! ways. Thus the total number of seating arrangement so that spouces sit next to each other = 3! × 2! × 2! × 2! = 48. • Again, if three ladies sit together, then necessarily three men must sit together. Thus, ladies and men can be arranged altogether among themselves in 2! ways. Therefore, the total number of ways where ladies sit together is 3! × 3! × 2! = 144.
- 19. • Combination: Picking a team of 3 people from a group of 10. C(10,3)=10!/(7! 3!)=10 9 8/(3 2 1)=120. ∗ ∗ ∗ ∗ ∗ • Permutation: Picking a President, VP and Waterboy from a group of 10. P(10,3)=10!/7!=10 9 8=720. ∗ ∗ • Combination: Choosing 3 desserts from a menu of 10. C(10,3) = 120. • Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.
- 20. Finding the Number of Permutations of n Non-Distinct Objects • If there are n elements in a set and r1, r2, r3, r4, and so on till rk that are all alike and indistinguishable, then the number of possible arrangements of the same set is given as the following: • n! / (r1! r2! r3! r4! …. rk!) • Find the number of rearrangements of the letters in the word DISTINCT. • Solution • There are 8 letters. Both I and T are repeated 2 times. Substitute n=8,r1 =2, • andr2 =2 into the formula. 8!/2!2!=10,080
- 21. • How many different “words” can we create by reordering “SUCCESS”? • A: 7 letters, including 3 S’s and 2 C’s with the rest unique: 7!/(3! 2!) ∙ • A classic permutation is linear while a circular permutation measures the possible arrangements of items in a circle. • The number of permutations of n elements in a circle is (n 1)! − • In how many ways can 6 people be seated at a round table?
- 22. Find the number of ways in which 5 people A, B, C, D, and E can be seated at a round table, such that (i) A and B always sit together. (ii) C and D never sit together. Solution (i) If we wish to seat A and B together in all arrangements, we can consider these two as one unit, along with 3 others. So, effectively we’ve to arrange 4 people in a circle, the number of ways being (4 – 1)! or 6.