Sign in

Libre University uses your GitHub account. Signing in is only needed to sit a final test, so the score is kept on your profile.

Inclusion and exclusion

1.[1p]

How many integers from 1 to 1000 are divisible by 3 or by 7?

CorrectNot quite: 428

2.[3p]

How many integers from 1 to 1000 are divisible by none of 2, 3, 5 and 7?

CorrectNot quite: 228

3.[1p]

An element lies in exactly four of the sets A1,…,An. How many times is it counted in S2, the sum of the sizes of all pairwise intersections?

CorrectNot quite: 6

4.[2p]

How many onto functions are there from a set of six elements to a set of three elements?

CorrectNot quite: 540

5.[2p]

Seven letters are put into seven addressed envelopes so that no letter is in its own envelope. In how many ways can this be done?

CorrectNot quite: 1854

6.[2p]

Six guests get their hats back at random. What is the probability that nobody gets their own hat? Give it to four decimal places.

CorrectNot quite: 0.3681

7.[2p]

Compute φ(504), where 504=23×32×7.

CorrectNot quite: 144

8.[3p]

Match each count to the sets Ai whose union the sieve removes.

  • Onto functions

  • Derangements

  • Euler's totient of n

  • the permutations that fix a given element

  • the integers up to n divisible by a given prime divisor of n

  • the functions that miss a given element of the target

Show the answer

Onto functions: the functions that miss a given element of the target Derangements: the permutations that fix a given element Euler's totient of n: the integers up to n divisible by a given prime divisor of n

9.[3p]

Which of these statements are correct?

Select all that apply

Correct
Correct
Correct
The answer is: The proof of the general formula uses the fact that the alternating sum of a row of Pascal's triangle is zero, For every $n \ge 1$, $D_n$ is $n!/e$ rounded to the nearest integer, $\varphi(n) = n\left(1 - \frac{1}{p_1}\right)\cdots\left(1 - \frac{1}{p_r}\right)$, where $p_1, \ldots, p_r$ are the distinct primes dividing $n$
The answer is: The proof of the general formula uses the fact that the alternating sum of a row of Pascal's triangle is zero, For every $n \ge 1$, $D_n$ is $n!/e$ rounded to the nearest integer, $\varphi(n) = n\left(1 - \frac{1}{p_1}\right)\cdots\left(1 - \frac{1}{p_r}\right)$, where $p_1, \ldots, p_r$ are the distinct primes dividing $n$
The answer is: The proof of the general formula uses the fact that the alternating sum of a row of Pascal's triangle is zero, For every $n \ge 1$, $D_n$ is $n!/e$ rounded to the nearest integer, $\varphi(n) = n\left(1 - \frac{1}{p_1}\right)\cdots\left(1 - \frac{1}{p_r}\right)$, where $p_1, \ldots, p_r$ are the distinct primes dividing $n$