Probability with Permutations and Combinations

Lesson 13-2: Probability with Permutations and Combinations

Five-Minute Check (over Lesson 13-1)

  • Question 1: Which is not part of the sample space for the following situation? George can eat at two different restaurants on his college campus. He has an hour break at 11:00 and at 1:00. Options: A. R1, 11 B. R1, R2 C. R2, 1 D. R2, 11
  • Question 2: Which is not part of the sample space for the following situation? An editor has two writers available to write a story. They can either write a factual piece or an editorial. Options: A. W1, E B. W2, E C. E, F D. W1, F
  • Question 3: Find the number of possible outcomes for the situation. When choosing a cell phone, Terrence has 4 color choices and 5 additional options. Options: A. 9 B. 15 C. 20 D. 40
  • Question 4: Find the number of possible outcomes for the situation. In a cafeteria there are 4 choices for a main dish, 4 choices for a side dish, 5 choices of drinks, and 2 choices for dessert. Options: A. 40 B. 120 C. 150 D. 160
  • Question 5: For her birthday, Trina received a new wardrobe consisting of 6 shirts, 4 pairs of pants, 2 skirts, and 3 pairs of shoes. How many new outfits can she make? Options: A. 144 B. 130 C. 94 D. 72

Targeted TEKS

  • G.13(A): Develop strategies to use permutations and combinations to solve contextual problems.
  • Mathematical Processes: G.1(A), G.1(E)

Then/Now

  • Then: You used the Fundamental Counting Principle.
  • Now:
    • Use permutations with probability.
    • Use combinations with probability.

New Vocabulary

  • Permutation
  • Factorial
  • Circular permutation
  • Combination

Key Concept: Factorial

  • Words: The factorial of a positive integer nn, written n!n!, is the product of the positive integers less than or equal to nn.
  • Symbols: n!=n⋅(n−1)⋅(n−2)⋅…⋅2⋅1n! = n \cdot (n-1) \cdot (n-2) \cdot … \cdot 2 \cdot 1, where 0!=10! = 1

Example 1: Probability and Permutations of n Objects

  • Problem: Eli and Mia, along with 30 other people, sign up to audition for a talent show. Contestants are called at random to perform for the judges. What is the probability that Eli will be called to perform first and Mia will be called second?
  • Step 1: Find the number of possible outcomes in the sample space. This is the number of permutations of the order of the 32 contestants, or 32!32!.
  • Step 2: Find the number of favorable outcomes. This is the number of permutations of the other contestants given that Eli is first and Mia is second, which is (32–2)!(32 – 2)! or 30!30!.
  • Step 3: Calculate the probability.
    • number of favorable outcomesnumber of possible outcomes=30!32!=30!32⋅31⋅30!=132⋅31=1992\frac{\text{number of favorable outcomes}}{\text{number of possible outcomes}} = \frac{30!}{32!} = \frac{30!}{32 \cdot 31 \cdot 30!} = \frac{1}{32 \cdot 31} = \frac{1}{992}
  • Answer: 1992\frac{1}{992}

Check Point

  • Hila, Anisa, and Brant are in a lottery drawing for housing with 40 other students to choose their dorm rooms. If the students are chosen in random order, what is the probability that Hila is chosen first, Anisa second, and Brant third?

Key Concept: Permutations

  • Symbols: The number of permutations of nn distinct objects taken rr at a time is denoted by nP<em>rnP<em>r and is given by nP</em>r=n!(n−r)!nP</em>r = \frac{n!}{(n-r)!}.
  • Example: The number of permutations of 5 objects taken 2 at a time is 5P2=5!(5−2)!=5⋅4⋅3!3!=5⋅4=205P_2 = \frac{5!}{(5-2)!} = \frac{5 \cdot 4 \cdot 3!}{3!} = 5 \cdot 4 = 20.

Example 2: Probability and nPr

  • Problem: There are 12 puppies for sale at the local pet shop. Four are brown, four are black, three are spotted, and one is white. What is the probability that all the brown puppies will be sold first?
  • Step 1: Since the order that the puppies are sold is important, this problem relates to permutation. The number of possible outcomes in the sample space is the number of permutations of 12 puppies taken 4 at a time which is <em>12P</em>4=12!(12−4)!=12!8!=12⋅11⋅10⋅9=11,880<em>{12}P</em>4 = \frac{12!}{(12-4)!} = \frac{12!}{8!} = 12 \cdot 11 \cdot 10 \cdot 9 = 11,880.
  • Step 2: The number of favorable outcomes is the number of permutations of the 4 brown puppies in their specific positions. This is 4!=4⋅3⋅2⋅1=244! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 favorable outcomes.
  • Step 3: So the probability of the four brown puppies being sold first is 2411,880=1495\frac{24}{11,880} = \frac{1}{495}.
  • Answer: 1495\frac{1}{495}

Check Point

  • There are 24 people in a hula-hoop contest. Five of them are part of the Garcia family. If everyone in the contest is equally as good at hula-hooping, what is the probability that the Garcia family finishes in the top five spots?

Key Concept: Permutations with Repetition

  • The number of distinguishable permutations of nn objects in which one object is repeated r<em>1r<em>1 times, another is repeated r</em>2r</em>2 times, and so on, is n!r<em>1!⋅r</em>2!⋅…⋅rk!\frac{n!}{r<em>1! \cdot r</em>2! \cdot … \cdot r_k!}.

Example 3: Probability and Permutations with Repetition

  • Problem: A box of floor tiles contains 5 blue (bl) tiles, 2 gold (gd) tiles, and 2 green (gr) tiles in random order. The desired pattern is bl, gd, bl, gr, bl, gd, bl, gr, and bl. If you selected a permutation of these tiles at random, what is the probability that they would be chosen in the correct sequence?
  • Step 1: There is a total of 9 tiles. Of these tiles, blue occurs 5 times, gold occurs 2 times, and green occurs 2 times. So the number of distinguishable permutations of these tiles is 9!5!⋅2!⋅2!=378\frac{9!}{5! \cdot 2! \cdot 2!} = 378.
  • Step 2: There is only one favorable arrangement— bl, gd, bl, gr, bl, gd, bl, gr, bl.
  • Step 3: The probability that a permutation of these tiles selected will be in the chosen sequence is 1378\frac{1}{378}.
  • Answer: 1378\frac{1}{378}

Check Point

  • A box of floor tiles contains 3 red (rd) tiles, 3 purple (pr) tiles, and 2 orange (or) tiles in random order. The desired pattern is rd, rd, pr, pr, or, rd, pr, and or. If you selected a permutation of these tiles at random, what is the probability that they would be chosen in the correct sequence?

Key Concept: Circular Permutations

  • The number of distinguishable permutations of nn objects arranged in a circle with no fixed reference point is (n−1)!(n - 1)!.

Example 4: Probability and Circular Permutations

  • A. Seating: If 8 students sit at random in the circle of chairs, what is the probability that the students sit in a specific arrangement?

    • Since there is no fixed reference point, this is a circular permutation. So there are (8–1)!=7!=5040(8 – 1)! = 7! = 5040 distinguishable permutations of the way the students can sit.
    • Probability=15040\text{Probability} = \frac{1}{5040}

  • B. Crayons: You purchase a box of 8 crayons. If the crayons are packaged in random order, what is the probability that the crayon on the far left is red?

    • Since the crayons are packaged in a row, instead of a circle with no fixed reference point, this is a linear permutation. In that case, since there are 8 positions and 1 red crayon, the probability that the crayon on the far left is red is 18\frac{1}{8}.

Check Point

  • A. Table Settings: If for a birthday party there are 5 people having cake, and there are 5 different colored plates, what is the probability that if chosen at random the plates will be displayed as seen in the order at the right?
  • B. Construction: A home builder is constructing 6 different models of homes on a major cross street, 5 of which are 2-floored homes, and only 1 home that is 1 floor. If built at random, what is the possibility the 1-floored home will be on the 1st plot of land?

Key Concept: Combinations

  • Symbols: The number of combinations of nn distinct objects taken rr at a time is denoted by nC<em>rnC<em>r and is given by nC</em>r=n!(n−r)!⋅r!nC</em>r = \frac{n!}{(n - r)! \cdot r!}.
  • Example: The number of combinations of 8 objects taken 3 at a time is 8C3=8!(8−3)!⋅3!=8!5!⋅3!=8⋅7⋅6⋅5!5!⋅3⋅2⋅1=8⋅7⋅66=568C_3 = \frac{8!}{(8 - 3)! \cdot 3!} = \frac{8!}{5! \cdot 3!} = \frac{8 \cdot 7 \cdot 6 \cdot 5!}{5! \cdot 3 \cdot 2 \cdot 1} = \frac{8 \cdot 7 \cdot 6}{6} = 56.

Example 5: Probability and nCr

  • A set of alphabet magnets are placed in a bag. If 5 magnets are drawn from the bag at random, what is the probability that they will be the letters a, e, i, o, and u?
  • Step 1: Since the order in which the magnets are chosen does not matter, the number of possible outcomes in the sample space is the number of combinations of 26 letters taken 5 at a time, <em>26C</em>5=26!(26−5)!5!=26!21!5!=65,780<em>{26}C</em>5 = \frac{26!}{(26-5)!5!} = \frac{26!}{21!5!} = 65,780.
  • Step 2: There is only one favorable outcome that all 5 letters are a, e, i, o, and u. The order in which they are chosen is not important.
  • Step 3: So, the probability of just getting a, e, i, o, and u is 165,780\frac{1}{65,780}.
  • Answer: 165,780\frac{1}{65,780}

Check Point

  • A set of alphabet magnets are placed in a bag. If 4 magnets are drawn from the bag at random, what is the probability that they will be the letters m, a, t, and h?