Archive Notes

FC1 Week 7 SGT

Properties1
tagsmodule4ccs1fc1

4 min read

1

  1. There are possible arrangements per question. There are total questions. Hence different ways to solve the test.
  2. There are different ways to solve each question, select one of the answers or none of them. Hence different ways to solve the test.
  3. This adds more different ways per question. Hence different ways to solve the test.

2

  1. By cases:

    • If all jobs are assigned to one person, there are ways.
    • If one gets and one gets , we can consider how we can pick out of employees, in this case repetition is not allowed but order does matter (e.g. AB or BA, A is assigned 2 or B is assinged 2), there are ways.
    • If three employees get job each, we can consider as before but instead looking at picking out of employees. There are ways.

    Hence, there are different ways we could assign these jobs.

  2. hence (order matters, repetition not allowed) different ways to assign jobs.

  2. choose

3

Order matters, repetitions not allowed. . Hence, .

Total is $5$ letters.
We have:
- 4 Es
- 1 V
- 2 Rs
- 1 G
- 1 N
 
We want to choose $4$ places to assign to E: $\begin{pmatrix} 9 \\ 4 \end{pmatrix}$.
We want to choose $2$ places to assign to R: $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$
For the remaining places: $3!$
 
Hence we find $\begin{pmatrix} 9 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 2 \end{pmatrix} \cdot 3! = 7560$.

4

so that there are at least numbers whose sum or difference is divisible by .

If you have different natural numbers, each of which whose sum or difference are not divisible by , such as the set . Then we can add any new unique number whose sum or difference will be divisible by . e.g. take , add any number, e.g. , the difference of and is divisible by .

You can also think of it as columns.

01..999

An element must be offset by from one of these columns at any one time. So if we have elements, either all columns are filled so any new elements will be offset by OR columns are filled more than once, but since the difference between subsequent rows is , but in this case we’ve already satisified the statement.

Generalised pigeonhole principle

The pigeonhole principle states that if objects are placed into boxes, then there is at least one box containing at least $\lceil n/k \rceil$ objects.

Link to original

title: Incorrect.
 
$$
	n = 1000k + r
$$
 
The sum ($n_1 + n_1$) is divisible by $1000$ in the cases where $r_1 + r_2 = 1000$ or $r_1 + r_2 = 0$.
The difference is divisible by $1000$ in cases where $r_1 = r_2$.
 
For any given $n$, $r$ can be any value from $0..999$.

5

To get a sum of , we have the combinations:

  • ;
  • ;
  • ;
  • ;
  • ;
  • ;

Hence, probability of sum of is:

6

title: Correct answers.

7

Let be that it is successful. Let be that it was predicted to be successful.

We know that:

Graph View