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Binomial coefficients

1.[2p]

Find the coefficient of x3 in (2x+1)6.

CorrectNot quite: 160

2.[2p]

Find the coefficient of x2y3 in (x-2y)5.

CorrectNot quite: -80

3.[1p]

What is (100)+(101)+(102)+⋯+(1010)?

CorrectNot quite: 1024

4.[2p]

Compute (60)2+(61)2+⋯+(66)2.

CorrectNot quite: 924

5.[3p]

A panel of four is chosen from five chemists and six biologists. How many panels include at least two chemists?

CorrectNot quite: 215

6.[2p]

Compute (22)+(32)+(42)+⋯+(92).

CorrectNot quite: 120

7.[2p]

Which double count proves k(nk)=n(n-1k-1)?

Correct
The answer is: Committees of $k$ from $n$ people with one member marked as chair, counted committee first and then chair first
The answer is: Committees of $k$ from $n$ people with one member marked as chair, counted committee first and then chair first
The answer is: Committees of $k$ from $n$ people with one member marked as chair, counted committee first and then chair first

8.[3p]

Match each identity to the way its set is split in the counting proof.

  • Pascal's rule

  • Sum of a row

  • Vandermonde's identity

  • Hockey stick

  • committees split by how many members come from the first group

  • subsets split by their size

  • subsets split by their largest element

  • committees split by whether one fixed person is on them

Show the answer

Pascal's rule: committees split by whether one fixed person is on them Sum of a row: subsets split by their size Vandermonde's identity: committees split by how many members come from the first group Hockey stick: subsets split by their largest element

9.[3p]

Which of these statements are correct?

Select all that apply

Correct
Correct
Correct
The answer is: A nonempty finite set has as many subsets of even size as of odd size, $\binom{n}{k} = \binom{n}{n-k}$, because sending a subset to its complement is a bijection, $\sum_{k=0}^{n} k\binom{n}{k} = n \cdot 2^{n-1}$
The answer is: A nonempty finite set has as many subsets of even size as of odd size, $\binom{n}{k} = \binom{n}{n-k}$, because sending a subset to its complement is a bijection, $\sum_{k=0}^{n} k\binom{n}{k} = n \cdot 2^{n-1}$