permutation and combination (1 Viewer)

[manscux]

hey guys help with this question

three identical pens and four identical pencils are arranged in a row. if only 5 writing implements are used how many arrangements are possible

 

[Magical Kebab]

7p5?
Might be wrong lol.

 

[manscux]

yah thats what i had..... nope the answer is 25... please show working

7p5?
Might be wrong lol.

 

[Magical Kebab]

yah thats what i had..... nope the answer is 25... please show working

u sure its 25?

 

[manscux]

u sure its 25?

yes

 

[Timske]

7+6+5+4+3

So you have 7 pens/pencils in total now you want to choose 5 of either.

_ _ _ _ _
First choice you can pick from 7,
Second you can pick from 6,
Third you can pick from 5
Fourth you can pick 4
Fifth you can pick 3

You shouldn't have given the answer away because the solution can be wrong yet it can still produce the same answer.

 

[Magical Kebab]

7+6+5+4+3

There u have it.

 

[RealiseNothing]

Let pens = A and pencils = B

We have 3 A's and 4 B's, and want to arrange 5 of them. So we have have (without regard to order):

BBBBA, BBBAA, and BBAAA.

Now let's take order into consideration:

BBBBA = 5 ways to arrange.

BBBAA = 10 ways to arrange.

BBAAA = 10 ways to arrange (since it's same as BBBAA just swap A and B around).

Hence 10+10+5 = 25.

 

[RealiseNothing]

7+6+5+4+3

So you have 7 pens/pencils in total now you want to choose 5 of either.

_ _ _ _ _
First choice you can pick from 7,
Second you can pick from 6,
Third you can pick from 5
Fourth you can pick 4
Fifth you can pick 3

You shouldn't have given the answer away because the solution can be wrong yet it can still produce the same answer.

Are you sure this method is valid?

 

[manscux]

Hey, i got another one...its the same type but i'm not sure of the concept....
In how many ways can 6 different books be distributed between 2 students, provided that both students recieve at least one book?

 

[wilsondw]

Its a valid method....seen some of the teachers and students do a question this way

 

[deterministic]

Hey, i got another one...its the same type but i'm not sure of the concept....
In how many ways can 6 different books be distributed between 2 students, provided that both students recieve at least one book?

Number of ways with both students with at least 1 book = Total number of ways - Number of ways that one student has no books

Using that, we have:
Total number of ways= 2x2x2x2x2x2 = 2^6 (each book can go to 2 positions)
Number of ways where one student has no books =2

Hence Number of ways with both students with at least 1 book = 2^6-1