C960 Discrete Mathematics II

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Free C960 Discrete Mathematics II Questions

1.

What is the expected number of coin flips until you get two heads in a row?

  • 4

  • 6

  • 8

  • 10

Explanation

To find the expected number of flips to get two consecutive heads, we can use a state-based approach. Let E0​ be the expected flips starting from no heads, and E1​ the expected flips starting with one head.


2.

In a single-elimination tournament with 16 teams, how many games are played until a champion is determined?

  • 15

  • 16

  • 8

  • 31

Explanation

In a single-elimination tournament, each game eliminates exactly one team. Starting with 16 teams, we need to eliminate 15 teams to determine a champion. Since each game eliminates one team, the total number of games played is equal to the number of teams eliminated, which is 16−1=15. This formula holds for any single-elimination tournament: total games = total teams − 1.


3.

How many ways can 7 people sit around a circular table if two specific people must sit together?

  • 240

  • 360

  • 480

  • 720

Explanation

For circular arrangements, we usually fix one seat to account for rotations, giving (n−1)! arrangements for nnn people.

Here, there are 7 people with two specific people required to sit together. Treat these two as a single "block," reducing the problem to 6 objects (the block + 5 other people). The number of circular arrangements is:

6-1!=5!=120

 


4.

Which of the following languages requires a linear bounded automaton (LBA)?

  • The set of palindromes over {0,1}

  • The set of strings of the form anbncn

  • The set of all binary strings

  • The set of all strings with an even number of 0s

Explanation

A linear bounded automaton (LBA) is a type of Turing machine whose tape usage is limited to a linear function of the input length. LBAs recognize context-sensitive languages, which are strictly more powerful than context-free languages but less powerful than general Turing machines.

Palindromes over {0,1} are context-free, so a pushdown automaton can handle them.

Strings of the form anbncn are context-sensitive but not context-free, so they require an LBA.

 


5.

What is the coefficient of x⁸ in the expansion of (x + 2 + x⁻¹)⁸?

  • 1

  • 70

  • 56

  • 64

Explanation


6.

Using Fermat’s Little Theorem, compute 3100mod 7.

  • 1

  • 2

  • 3

  • 4

Explanation

Fermat’s Little Theorem states that if ppp is a prime number and aaa is an integer not divisible by p, then ap-1−1≡1(mod p).


7.

How many 5-digit numbers have digits that sum to 20?

  • 106

  • 121

  • 126

  • 210

Explanation


8.

In a group of 30 people, what is the minimum number that guarantees at least 4 people share the same birth month?

  • 37

  • 10

  • 13

  • 4

Explanation

This is an application of the Pigeonhole Principle. There are 12 months (pigeonholes) and 30 people (pigeons). To guarantee at least 4 people in the same month, consider the worst-case scenario where no month has 4 people: each month could have at most 3 people.

12 months×3 people per month=36

Wait, we need the minimum number that guarantees 4 people in a month. In the worst case, if each month has at most 3 people, then 12 × 3 = 36 people could avoid having 4 in a month. But we only have 30 people, which is less than 36.

To guarantee 4 people in a month, the minimum number of people needed is 3 × 12 + 1 = 37.

Since we have only 30 people, we cannot guarantee 4 in a month.

If the question is asking more generally: the formula is:

Minimum number to guarantee k in a box=(k−1)⋅n+1

Here, k=4, n=12:

(4−1)⋅12+1=37


9.

How many Hamiltonian cycles are there in the complete graph K₅?

  • 12

  • 24

  • 60

  • 120

Explanation

A Hamiltonian cycle in a graph visits each vertex exactly once and returns to the starting vertex. In a complete graph kn, every vertex is connected to every other vertex.

 


10.

How many ways can you arrange the letters in “MISSISSIPPI”?

  • 34,650

  • 36,300

  • 37,800

  • 40,000

Explanation

The word MISSISSIPPI has 11 letters in total. The counts of each letter are:

M = 1

I = 4

S = 4

P = 2

The total number of distinct arrangements is given by dividing the factorial of the total letters by the factorial of the frequency of each repeated letter:

Number of arrangements=11!1!.4!.4!.2!



11!=39916800;4!=24;2!=2

so, 3991680024×24×2=399168001152=34650. 

 


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