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Permutation with repetition

When order matters and an object can be chosen more than once then the number of

permutations is:

n r

where n is the number of objects from which you can choose and r is the number to be chosen.

For example, if you have the letters A, B, C, and D and you wish to discover the number of ways of arranging them in three letter patterns (trigrams) you find that there are 4 3 or 64 ways. This is because for the first slot you can choose any of the four values, for the second slot you can choose any of the four, and for the final slot you can choose any of the four letters. Multiplying them together gives the total.

Applications

The binomial theorem

In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. Its simplest version reads

( x + y ) n = k = 0 n n k x k y n - k

Whenever n is a positive integer, the numbers

n k = n ! k ! ( n - k ) !

are the binomial coefficients (the coefficients in front of the powers).

For example, here are the cases n = 2, n = 3 and n = 4:

( x + y ) 2 = x 2 + 2 x y + y 2 ( x + y ) 3 = x 3 + 3 x 2 y + 3 x y 2 + y 3 ( x + y ) 4 = x 4 + 4 x 3 y + 6 x 2 y 2 + 4 x y 3 + y 4

The coefficients form a triangle, where each number is the sum of the two numbers above it:

This formula, and the triangular arrangement of the binomial coefficients, are often attributed to Blaise Pascal who described them in the 17th century. It was, however, known to the Chinese mathematician Yang Hui in the 13th century, the earlier Persian mathematician Omar Khayyám in the 11th century, and the even earlier Indian mathematician Pingala in the 3rd century BC.

The number plate on a car consists of any 3 letters of the alphabet (excluding the vowels and 'Q'), followed by any 3 digits (0 to 9). For a car chosen at random, what is the probability that the number plate starts with a 'Y' and ends with an odd digit?

  1. The number plate starts with a 'Y', so there is only 1 choice for the first letter, and ends with an even digit, so there are 5 choices for the last digit (1, 3, 5, 7, 9).

  2. Use the counting principle. For each of the other letters, there are 20 possible choices (26 in the alphabet, minus 5 vowels and 'Q') and 10 possible choices for each of the other digits.

    Number of events = 1 × 20 × 20 × 10 × 10 × 5 = 200 000

  3. Use the counting principle. This time, the first letter and last digit can be anything.

    Total number of choices = 20 × 20 × 20 × 10 × 10 × 10 = 8 000 000

  4. The probability is the number of events we are counting, divided by the total number of choices.

    Probability = 200 000 8 000 000 = 1 40 = 0 , 025

Show that

n ! ( n - 1 ) ! = n
  1. Method 1: Expand the factorial notation.

    n ! ( n - 1 ) ! = n × ( n - 1 ) × ( n - 2 ) × . . . × 2 × 1 ( n - 1 ) × ( n - 2 ) × . . . × 2 × 1

    Cancelling the common factor of ( n - 1 ) × ( n - 2 ) × . . . × 2 × 1 on the top and bottom leaves n .

    So n ! ( n - 1 ) ! = n

  2. Method 2: We know that P ( n , r ) = n ! ( n - r ) ! is the number of permutations of r objects, taken from a pool of n objects. In this case, r = 1 . To choose 1 object from n objects, there are n choices.

    So n ! ( n - 1 ) ! = n

Exercises

  1. Tshepo and Sally go to a restaurant, where the menu is:
    Starter Main Course Dessert
    Chicken wings Beef burger Chocolate ice cream
    Mushroom soup Chicken burger Strawberry ice cream
    Greek salad Chicken curry Apple crumble
    Lamb curry Chocolate mousse
    Vegetable lasagne
    1. How many different combinations (of starter, main course, and dessert) can Tshepo have?
    2. Sally doesn't like chicken. How many different combinations can she have?
  2. Four coins are thrown, and the outcomes recorded. How many different ways are there of getting three heads? First write out the possibilities, and then use the formula for combinations.
  3. The answers in a multiple choice test can be A, B, C, D, or E. In a test of 12 questions, how many different ways are there of answering the test?
  4. A girl has 4 dresses, 2 necklaces, and 3 handbags.
    1. How many different choices of outfit (dress, necklace and handbag) does she have?
    2. She now buys two pairs of shoes. How many choices of outfit (dress, necklace, handbag and shoes) does she now have?
  5. In a soccer tournament of 9 teams, every team plays every other team.
    1. How many matches are there in the tournament?
    2. If there are 5 boys' teams and 4 girls' teams, what is the probability that the first match will be played between 2 girls' teams?
  6. The letters of the word 'BLUE' are rearranged randomly. How many new words (a word is any combination of letters) can be made?
  7. The letters of the word 'CHEMISTRY' are arranged randomly to form a new word. What is the probability that the word will start and end with a vowel?
  8. There are 2 History classes, 5 Accounting classes, and 4 Mathematics classes at school. Luke wants to do all three subjects. How many possible combinations of classes are there?
  9. A school netball team has 8 members. How many ways are there to choose a captain, vice-captain, and reserve?
  10. A class has 15 boys and 10 girls. A debating team of 4 boys and 6 girls must be chosen. How many ways can this be done?
  11. A secret pin number is 3 characters long, and can use any digit (0 to 9) or any letter of the alphabet. Repeated characters are allowed. How many possible combinations are there?

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In biology, a pathogen (Greek: πάθος pathos "suffering", "passion" and -γενής -genēs "producer of") in the oldest and broadest sense, is anything that can produce disease. A pathogen may also be referred to as an infectious agent, or simply a germ. The term pathogen came into use in the 1880s.[1][2
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Source:  OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2
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