(EN) - NumberTheory_Combinatory1_EFQ2

COMBINATORY

PERMUTATION, VARIATION & COMBINATION

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Hi, mate! I'm Jake. Nice to meet you!

I’m ordering my grandma's personal library. I think you can be helpful.

I’ve found these old books, files, I don't know what they are. They don’t seem to follow a specific order. How many ways do you think I can sort them?

3

6

9

1

2

3

Are you sure? Try again!

Ups! Try again!

CONTINUE

Correct!

Looks simple, doesn't it? But what if we were increasing the number of files?

Would you use the same method to know all the ways you could sort them?

CONTINUE

Lets see what happened:

You had to order 3 different books, in many possible order, and you can't repeat them because they are unique.

So...

1. You use all the elements.

2. Order matters.

3. There's no repetitions.

THIS IS CALLED

PERMUTATION

Pn=n!

CONTINUE

P3 = 3! = 3*2*1 = 6

PERMUTATION

Pn=n!

3 books

6 different orders

P4 = 4! = 4*3*2*1 = 24

P5 = 5! = 5*4*3*2*1 = 120

...

Could you imagine?!

Continue

Hi, mate! I'm Sally. Nice to meet you!

WRONG PASSWORD

I'm very worried... I've forgotten my email's password, and I need it as soon as possible. May you help me?

I only remember that it's a 6 number password, and every number is different. I'm in a hurry, so I'm thinking in start testing number by number until I find it. Do you think that's a good idea?

Of course! Start trying now!

I don't think so... Just give up

Nop, that's not! Let me help you.

*

*

*

*

*

*

INCORRECT.

INCORRECT.

CONTINUE

Correct!

Lets see what we can do...

CONTINUE

Lets see what happened:

Sally had to remember 6 different digits, from 10 numbers from 0 to 9, without repeating any.

So...

1. You don't use all the elements.

2. Order matters.

3. There's no repetitions.

THIS IS CALLED

VARIATION

Vm,n = m!/(m-n)!

CONTINUE

V10,6 = 10! / (10-6)!

VARIATION

Vm,n = m!/(m-n)!

10 numbers, 6 digits

151.200 different orders

V10,6 = 10*9*8*7*6*5 = 151.200

And just one is correct!

Continue

Hi, mate! I'm Louisa. May you help me?

I just have won three tickets to go see the premiere of a movie tonight, but I don't know who to go with.

I have three friends who want to go, but I can only take two and I don't know who to choose... How many options do I have?

2

3

6

Are you sure? Try again!

Ups! Try again.

CONTINUE

Correct!

Looks simple, doesn't it? But what if we were increasing the number of friends?Would you choose the same method to solve this?

CONTINUE

Lets see what happened:

Luisa has 3 tickets, 1 for herself and 2 for her friends. She has 3 friends, so she needs to rule one out.

So...

1. You don't use all the elements.

2. Order doesn't matter.

3. There's no repetitions.

THIS IS CALLED

COMBINATION

Cm,n = m!/(n!*(m-n)!)

CONTINUE

C3,2 = 3! / 2! = 3

COMBINATION

Cm,n = m!/(n!*(m-n)!)

3 friends, 2 tickets

3 different couples of friends

C7,4 = 7!/(4!*3!) = 35

Etc.

...

CONTINUE

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READING THE PROBLEM: WHAT TO LOOK FOR

WELL DONE!