Free Online Flashcard Deck

07 Advanced Counting Methods Free Online FlashCards

Study 07 Advanced Counting Methods with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

What does (nk)\binom{n}{k} count?

Back

a binomial coefficient

(nk)\binom{n}{k}

counts the unordered kk-element subsets of an nn-element set, where (nk)=n!k!(n−k)!\binom{n}{k}=\frac{n!}{k!(n-k)!}.

02
Front

State the symmetry identity for binomial coefficients.

Back

Symmetry gives (nk)=(nn−k)\binom{n}{k}=\binom{n}{n-k}, because choosing the selected objects is equivalent to choosing the objects left out.

03
Front

What is Pascal's identity?

Back

Pascal's identity is (nk)=(n−1k−1)+(n−1k)\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}. It separates selections according to whether a distinguished object is chosen.

04
Front

How many selections with repetition are possible?

Back

The number of ways to select kk objects from nn types when repetition is allowed is (n+k−1k)\binom{n+k-1}{k}.

05
Front

State the binomial theorem.

Back

The binomial theorem states (x+y)n=∑k=0n(nk)xn−kyk(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k for nonnegative integer nn.

06
Front

How do you extract a coefficient from (x+y)n(x+y)^n?

Back

The coefficient of xn−kykx^{n-k}y^k in (x+y)n(x+y)^n is (nk)\binom{n}{k}. For example, the coefficient of x5y3x^5y^3 in (x+y)8(x+y)^8 is (83)=56\binom{8}{3}=56.

07
Front

What is inclusion-exclusion for two sets?

Back

For two finite sets, ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣|A\cup B|=|A|+|B|-|A\cap B|. The intersection is subtracted because it was counted twice.

08
Front

What sign pattern does three-set inclusion-exclusion use?

Back

For three sets, add single-set sizes, subtract pairwise intersections, then add the triple intersection: ∣A∪B∪C∣=∑∣A∣−∑∣A∩B∣+∣A∩B∩C∣|A\cup B\cup C|=\sum|A|-\sum|A\cap B|+|A\cap B\cap C|.

09
Front

How does inclusion-exclusion count objects avoiding all restrictions?

Back

If AiA_i is the set violating restriction ii, then valid objects number ∣S∣−∣⋃iAi∣|S|-|\bigcup_i A_i|, equivalently ∣⋂iAic∣|\bigcap_i A_i^c|.

10
Front

What formula counts derangements?

Back

The number of derangements of nn objects is Dn=n!∑k=0n(−1)kk!D_n=n!\sum_{k=0}^{n}\frac{(-1)^k}{k!}, and approximately n!e\frac{n!}{e}.

11
Front

What makes a recurrence-based counting argument complete?

Back

A complete recurrence description needs initial conditions, a recurrence relation, and a justification based on an exhaustive, disjoint case partition.

12
Front

What recurrence counts binary strings with no consecutive 1s?

Back

For binary strings with no consecutive 1s, an=an−1+an−2a_n=a_{n-1}+a_{n-2}, with a0=1a_0=1 and a1=2a_1=2. The cases end in 0 or in 01.