What are the principles of counting?

The fundamental counting principle states that if there are p ways to do one thing, and q ways to do another thing, then there are p × q ways to do both things.

Example 1:

Suppose you have 3 shirts (call them A , B , and C ), and 4 pairs of pants (call them w , x , y , and z ). Then you have

3 × 4 = 12

possible outfits:

A w , A x , A y , A z B w , B x , B y , B z C w , C x , C y , C z

Example 2:

Suppose you roll a 6 -sided die and draw a card from a deck of 52 cards. There are 6 possible outcomes on the die, and 52 possible outcomes from the deck of cards. So, there are a total of

6 × 52 = 312

possible outcomes of the experiment.

The counting principle can be extended to situations where you have more than 2 choices. For instance, if there are p ways to do one thing, q ways to a second thing, and r ways to do a third thing, then there are p × q × r ways to do all three things.

Lily is trying to decide what to wear. She has shirts in the following colors: red, purple, and blue, and she has pants in the following colors: black and white. How many different outfits can Lily choose from (assuming she selects one shirt and one pair of pants)?

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We know from the definition of the rule of product that if there are nn options for doing one thing (like choosing a shirt), and mm options for doing another thing (like choosing a pair of pants), then there are n×mn \times m total combinations we can choose from. In this case, there are 33 options for choosing a shirt, and there are 22 options for choosing pants. Thus, there are 3×2=63 \times 2 = 6 total options.

Here is a table where each row represents a possible outfit.

Shirt	Pants
Red	Black
Blue	Black
Purple	Black
Red	White
Blue	White
Purple	White

As expected, there are 66 possible combinations. □_\square

In the example above, there were two things to choose: a shirt and a pair of pants. However, the rule of product can extend to however many things to choose from. For example, if there are nn choices for a shirt, mm choices for a pair of pants, x x choices for a pair of shoes, and yy choices for a hat, the rule of product states that there are n×m×x×yn \times m \times x \times y total possible combinations.

Unknown 175 145000 15000 142500

You go to check out three books at the library, and you want one history book, one science book, and one fantasy book.

The library has 50 history books, 95 fantasy novels, and 30 books about science. How many combinations of books do you have to choose from?

There are 8 8 daily newspapers and 55 weekly magazines published in Chicago. If Colin wants to subscribe to exactly one daily newspaper and one weekly magazine, how many different choices does he have?

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Colin has 8×5 =408\times5=40 choices. □_\square

Calvin wants to go to Milwaukee. He can choose from 3 3 bus services or 22 train services to head from home to downtown Chicago. From there, he can choose from 2 bus services or 3 train services to head to Milwaukee. How many ways are there for him to get to Milwaukee?

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Since Calvin can either take a bus or a train downtown , he has 3+2=5 3+2 =5 ways to head downtown (Rule of sum). After which, he can either take a bus or a train to Milwaukee, and hence he has another 2+3=5 2+3=5 ways to head to Milwaukee (Rule of sum). Thus in total, he has 5×5=25 5 \times 5 = 25 ways to head from home to Milwaukee (Rule of product). □_\square

Six friends Andy, Bandy, Candy, Dandy, Endy, and Fandy want to sit in a row at the cinema. If there are only six seats available, how many ways can we seat these friends?

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For the first seat, we have a choice of any of the 6 friends. After seating the first person, for the second seat, we have a choice of any of the remaining 5 friends. After seating the second person, for the third seat, we have a choice of any of the remaining 4 friends. After seating the third person, for the fourth seat, we have a choice of any of the remaining 3 friends. After seating the fourth person, for the fifth seat, we have a choice of any of the remaining 2 friends. After seating the fifth person, for the sixth seat, we have a choice of only 1 of the remaining friends. Hence, by the rule of product, there are 6×5×4×3×2 ×1=720 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 ways to seat these 6 people. More generally, this problem is known as a permutation. There are n!=n×( n−1)×(n−2)×⋯×1 n! = n \times (n-1) \times (n-2) \times \cdots \times 1 ways to seat n n people in a row. □_\square

My toy piano keyboard has 7 distinct white notes: letters A-G in English alphabet. I'm going to create a melody by playing three random notes. I am not allowed to repeat any notes and the melody cannot be ended with E, F or G. How many different melodies can I play?

Examples:

• C G A is permitted.
• A F A isn't permitted because of repetition.
• A B E is not permitted because of last note rule.

How many positive divisors does 2000=2453 2000 = 2^4 5^3 have?

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Any positive divisor of 2000 must have the form 2a5b 2^a 5^b, where a a and b b are integers satisfying 0≤a≤4,0≤b≤3 0 \leq a \leq 4, 0 \leq b \leq 3. There are 5 possibilities for a a and 4 possibilities for b b, hence there are 5×4=20 5 \times 4 = 20 (rule of product) positive divisors of 2000 in all. □_\square

• Problem solving

• Rule of Sum
• Permutations

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