Free Online Flashcard Deck

2 Counting Techniques Free Online FlashCards

Study 2 Counting Techniques with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

What does the addition principle count?

Back

For mutually exclusive alternatives, add the numbers of choices: m+nm+n.

02
Front

How do you count overlapping categories?

Back

Subtract the overlap: ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣|A\cup B|=|A|+|B|-|A\cap B|.

03
Front

What does the multiplication principle count?

Back

Multiply the numbers of choices at successive stages: n1n2⋯nkn_1n_2\cdots n_k.

04
Front

What is a factorial?

Back

For a positive integer, n!=n(n−1)⋯2⋅1n!=n(n-1)\cdots2\cdot1; by definition, 0!=10!=1.

05
Front

What is a permutation?

Back

A permutation is an arrangement in which order matters. The count is P(n,r)=n!(n−r)!P(n,r)=\frac{n!}{(n-r)!}.

06
Front

How are arrangements with repeated objects counted?

Back

For repeated objects, divide by the factorial of each repetition count: n!n1!n2!⋯nk!\frac{n!}{n_1!n_2!\cdots n_k!}.

07
Front

What is a combination?

Back

A combination is an unordered selection. Its count is (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!}.

08
Front

How do you choose between permutations and combinations?

Back

Use a permutation when changing order creates a different outcome; use a combination when it does not.

09
Front

How are permutations and combinations related?

Back

The relationship is P(n,r)=(nr)r!P(n,r)=\binom{n}{r}r!: first choose the objects, then arrange them.

10
Front

What is the finite equally likely probability formula?

Back

For equally likely outcomes, P(A)=∣A∣∣S∣P(A)=\frac{|A|}{|S|}, where ∣A∣|A| is favorable outcomes and ∣S∣|S| is total outcomes.

11
Front

How many five-card hands are possible?

Back

The number of five-card hands is (525)=2,598,960\binom{52}{5}=2{,}598{,}960, because deal order does not matter.

12
Front

How is exactly two aces counted in a five-card hand?

Back

Count favorable hands as (42)(483)\binom{4}{2}\binom{48}{3}, then divide by (525)\binom{52}{5}.