Skills Beta

Factorials and Combinations

A factorial n!

Practice 3: Analyze Data

Part 1 · Hook

Why this matters

A student guesses on a 10-question quiz and gets 3 right. Which 3? It could be the first three, the last three, or any of many other sets. To find how likely 3 right answers is, you first need to count how many different sets of 3 questions there are. Factorials and combinations do that counting without listing every case.

Part 2 · Prerequisite check

Quick check before you start

1. Evaluate 6×52×1\frac{6 \times 5}{2 \times 1}.

  1. 15
  2. 30
  3. 12
  4. 3
Show the answer

Top: 30. Bottom: 2. 30/2 = 15.

  • Correct: 15:
  • 30:
  • 12:
  • 3:

2. A count X can be 0 to 10. Which values are "at least 9"?

  1. 9 and 10
  2. 10 only
  3. 0 to 9
  4. 9 only
Show the answer

At least 9 includes 9.

  • Correct: 9 and 10:
  • 10 only:
  • 0 to 9:
  • 9 only:

Part 3 · See it

See it first

All six teams of 2 that can be chosen from Ana, Ben, Cal and Dee: Ana and Ben, Ana and Cal, Ana and Dee, Ben and Cal, Ben and Dee, Cal and Dee. Below: 4C2 = 4! / (2! × 2!) = 24 / 4 = 6 teams.
All six teams of 2 from 4 people. A team is a set, so order does not matter: 4C2=6{}_4C_2 = 6. LevlPrep original diagram.

Part 4 · Step by step

How it works, step by step

  1. Line up n different itemsthere are n choices for the first place, n − 1 for the next, and so on: n!n! orders
  2. Choose k items in ordern×(n−1)×⋯n \times (n - 1) \times \cdots for k places, but each set of k appears k!k! times
  3. Divide out the ordersnCk{}_nC_k=n!k!(n−k)!= \frac{n!}{k!\,(n - k)!} counts the sets of k
  4. Choosing k to include is choosing n−kn - k to leave outnCk=nCn−k{}_nC_k = {}_nC_{n-k}, and nC0=nCn=1{}_nC_0 = {}_nC_n = 1

Part 5 · Key ideas

Key ideas

  • A factorial n!=n×(n−1)×⋯×2×1n! = n \times (n - 1) \times \cdots \times 2 \times 1 counts the orders of n different items; 0!=10! = 1.
  • A combination nCk{}_nC_k=(nk)= \binom{n}{k}=n!k!(n−k)!= \frac{n!}{k!\,(n - k)!} counts the ways to choose k of n items when order does not matter.
  • Ordered choices overcount: 10×9×8=72010 \times 9 \times 8 = 720 counts each set of 3 six times, so 10C3=120{}_{10}C_3 = 120.
  • On a calculator or in Bluebook, type 10C3{}_{10}C_3 as 10C3 or nCr(10, 3).

Part 6 · Misconception

A common mistake

The wrong idea: There are 4×3=124 \times 3 = 12 teams of 2 from 4 people.

What actually happens: 12 counts ordered pairs, so Ana with Ben is counted twice. Divide by 2!=22! = 2: 4C2=6{}_4C_2 = 6.

Part 7 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Type an expression for the number of ways to line up 5 different books on a shelf, and check its value.

Type the expression as on the exam, with the values substituted.

Show the answer

5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120 orders.

  • Answer: 5!5!

2. What is 4!4!?

  1. 16
  2. 24
  3. 10
  4. 4
Show the answer

n!=n×(n−1)×⋯×2×1n! = n \times (n - 1) \times \cdots \times 2 \times 1, so 4!=244! = 24.

  • 16: That is 424^2, not the factorial.
  • Correct: 24: Right: 4!=4×3×2×1=244! = 4 \times 3 \times 2 \times 1 = 24.
  • 10: That is 4+3+2+14 + 3 + 2 + 1; a factorial multiplies.
  • 4: A factorial multiplies 4 by every whole number below it.

3. A quiz has 10 questions. Type an expression for the number of ways to choose which 3 questions a student gets right.

Type the expression as on the exam, with the values substituted.

Show the answer

10C3{}_{10}C_3=10!3!7!= \frac{10!}{3!\,7!}=10×9×83×2×1= \frac{10 \times 9 \times 8}{3 \times 2 \times 1}=120= 120.

  • Answer: (103)\binom{10}{3}

4. How many different teams of 2 can be chosen from 4 people?

  1. 12
  2. 8
  3. 16
  4. 6
Show the answer

A team does not depend on order: 4C2=4×32×1=6{}_4C_2 = \frac{4 \times 3}{2 \times 1} = 6.

  • 12: That counts ordered pairs: Ana then Ben is the same team as Ben then Ana, so divide by 2.
  • 8: That is 4 × 2, which is not how choices combine.
  • 16: That is 424^2, which allows the same person twice and counts order.
  • Correct: 6: Right: 4C2=4!2!2!=6{}_4C_2 = \frac{4!}{2!\,2!} = 6.

5. Why does choosing 3 of 10 questions use a combination and not 10×9×810 \times 9 \times 8?

  1. The order in which the 3 questions are picked does not matter
  2. 10×9×810 \times 9 \times 8 is too large for a calculator
  3. Combinations are used when questions can be repeated
  4. 10×9×810 \times 9 \times 8 counts the questions answered wrong
Show the answer

Ordered choices overcount each set by the number of its orders. Dividing by 3!3! gives the number of sets: 7206=120\frac{720}{6} = 120.

  • Correct: The order in which the 3 questions are picked does not matter: Right: 10×9×8=72010 \times 9 \times 8 = 720 counts each set of 3 questions 3!=63! = 6 times, once for each order.
  • 10×9×810 \times 9 \times 8 is too large for a calculator: Calculators handle 720 easily; the problem is counting each set several times.
  • Combinations are used when questions can be repeated: A combination chooses different items, with no repeats.
  • 10×9×810 \times 9 \times 8 counts the questions answered wrong: It counts ordered choices of 3, not the wrong answers.

6. A pizza shop offers 6 toppings. How many different pizzas with exactly 3 different toppings are possible?

  1. 120
  2. 18
  3. 216
  4. 20
Show the answer

Order does not matter and toppings do not repeat: 6C3=20{}_6C_3 = 20.

  • 120: That counts ordered choices; the order of toppings on a pizza does not matter.
  • 18: That is 6 × 3, which is not a count of choices.
  • 216: That is 636^3, which allows the same topping three times.
  • Correct: 20: Right: 6C3{}_6C_3=6×5×43×2×1= \frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20= 20.

Part 8 · Summary

Summary

A factorial n!n! multiplies n by each whole number below it and counts the orders of n items (0!=10! = 1). A combination nCk{}_nC_k=n!k!(n−k)!= \frac{n!}{k!\,(n - k)!} counts the ways to choose k of n when order does not matter; it equals nCn−k{}_nC_{n-k}.

Part 9 · Up next

What comes next

Part 10 · Connections

Connections

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