MHB Distribute books to three people

[evinda]

Hey! ;)

I am looking at this exercise:

With how many ways can we distribute $21$ different books to the people $A,B \text{ and } C$,so that $A$ and $B$ get twice as much as $C$.
That's what I thought:
$x_1=\text{ the number of books that A gets }$
$x_2=\text{ the number of books that B gets }$
$x_3=\text{ the number of books that C gets }$

So,it is $x_1+x_2+x_3=21$ $(1)$.
$x_1+x_2=2 \cdot x_3$

Replacing $x_1+x_2=2 \cdot x_3$ at the relation $(1)$,we get $x_3=7$.
So, $C$ gets $\binom{21}{7}$ books.

$14$ books are left for $A$ and $B$ and the number of ways that they can get these books is equal to $14^{2}$.

So,in total,there are $\binom{21}{7} \cdot 14^{2}$ ways to distribute $21$ different books to the people $A,B \text{ and } C$,so that $A$ and $B$ get twice as much as $C$.

Is that what I have done right?? (Thinking)

 

[I like Serena]

Hey! ;)

I am looking at this exercise:

With how many ways can we distribute $21$ different books to the people $A,B \text{ and } C$,so that $A$ and $B$ get twice as much as $C$.
That's what I thought:
$x_1=\text{ the number of books that A gets }$
$x_2=\text{ the number of books that B gets }$
$x_3=\text{ the number of books that C gets }$

So,it is $x_1+x_2+x_3=21$ $(1)$.
$x_1+x_2=2 \cdot x_3$

Replacing $x_1+x_2=2 \cdot x_3$ at the relation $(1)$,we get $x_3=7$.
So, $C$ gets $\binom{21}{7}$ books.

$14$ books are left for $A$ and $B$ and the number of ways that they can get these books is equal to $14^{2}$.

So,in total,there are $\binom{21}{7} \cdot 14^{2}$ ways to distribute $21$ different books to the people $A,B \text{ and } C$,so that $A$ and $B$ get twice as much as $C$.

Is that what I have done right?? (Thinking)

Heya!

The wording of the question is a bit confusing.
Do $A$ and $B$ together get twice as much as $C$?
Or do $A$ and $B$ each get twice as much as $C$?
I would expect the last but perhaps I'm wrong.

Anyway, if we take the first meaning, which is what you did, the part about $C$ is good! :D

But... how did you find $14^2$?

 

[evinda]

Heya!

The wording of the question is a bit confusing.
Do $A$ and $B$ together get twice as much as $C$?
Or do $A$ and $B$ each get twice as much as $C$?
I would expect the last but perhaps I'm wrong.

Anyway, if we take the first meaning, which is what you did, the part about $C$ is good! :D

But... how did you find $14^2$?

It is meant that $A$ and $B$ together get twice as much as $C$!

I thought that $A$ and $B$ have both the possibility to get all of the $14$ books,but...now I think that it is wrong.. (Worried)(Thinking)

 

[I like Serena]

It is meant that $A$ and $B$ together get twice as much as $C$!

Good that we're clear on that! (Wink)

I thought that $A$ and $B$ have both the possibility to get all of the $14$ books,but...now I think that it is wrong.. (Worried)(Thinking)

Well... that does seem like a possibility... or rather 2 possibilities... (Thinking)

 

[evinda]

Good that we're clear on that! (Wink)

Well... that does seem like a possibility... or rather 2 possibilities... (Thinking)

So, $A$ and $B$ have both the possibility to get $14$ books,so there $2^{14}$ ways?? (Thinking)

 

[I like Serena]

So, $A$ and $B$ have both the possibility to get $14$ books,so there $2^{14}$ ways?? (Thinking)

That looks much better! (Emo)

Did you know that:
$$\binom n 0 + \binom n 1 + ... + \binom n n = 2^n$$

 

[evinda]

Did you know that:
$$\binom n 0 + \binom n 1 + ... + \binom n n = 2^n$$

Could you explain me why we can explain it,using this formula? I haven't understood it.. (Blush) (Thinking)

 

[I like Serena]

Could you explain me why we can explain it,using this formula? I haven't understood it.. (Blush) (Thinking)

$A$ has 1 way to get 0 books, meaning all 14 books go to $B$.
That is $\binom {14} 0 = 1$.

Then there are $\binom {14} 1 = 14$ ways that $A$ gets 1 book. The other 13 books go to $B$.

All in all, there are $\binom {14} 0 + \binom {14} 1 + ... + \binom {14} {14}$ ways that $A$ can get 0 up to 14 books, with the remainder going to $B$.

Alternatively, each book can either go to $A$ or to $B$.
That is 2 possibilities for each book.
That makes a total of $2^{14}$.
(Mmm)

 

[evinda]

$A$ has 1 way to get 0 books, meaning all 14 books go to $B$.
That is $\binom {14} 0 = 1$.

Then there are $\binom {14} 1 = 14$ ways that $A$ gets 1 book. The other 13 books go to $B$.

All in all, there are $\binom {14} 0 + \binom {14} 1 + ... + \binom {14} {14}$ ways that $A$ can get 0 up to 14 books, with the remainder going to $B$.

Alternatively, each book can either go to $A$ or to $B$.
That is 2 possibilities for each book.
That makes a total of $2^{14}$.
(Mmm)

I understand :) Thank you very much :D

 

[MarkFL]

Could you explain me why we can explain it,using this formula? I haven't understood it.. (Blush) (Thinking)

Hint: think of the binomial theorem...:D

 

[evinda]

Hint: think of the binomial theorem...:D

So,we use the formula $(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$ ?
We take $n=14$,or not? And what is then equal to $x$ and $y$ ?

 

[MarkFL]

So,we use the formula $(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$ ?
We take $n=14$,or not? And what is then equal to $x$ and $y$ ?

I was referring to your statement that you didn't understand the formula given by I like Serena:

$$\sum_{k=0}^n{n\choose k}=2^n$$

I meant for you to consider what the binomial theorem gives us for:

$$2^n=(1+1)^n$$

:D

 

[evinda]

I was referring to your statement that you didn't understand the formula given by I like Serena:

$$\sum_{k=0}^n{n\choose k}=2^n$$

I meant for you to consider what the binomial theorem gives us for:

$$2^n=(1+1)^n$$

:D

I have understood why the sum is equal to $2^n$,I hadn't understood why we take the sum to solve the exercise.. (Blush)