Combinatorics Word Problem: Choosing Hymns for a Sunday Service

[Petrus]

Hello MHB,
I got difficult understand this kind of questions, anyone got any tips or could explain, cause I find it really hard to understand when I do this kind of problem, this is an old problem from exam that I hopefully did translate well.
A priest needs to choose six hymns for Sunday's show. Hen has three hymnbooks, each with 25
hymns (thus 75 different hymns together). In how many ways can hen choose the hymns of
(a) all the hymns to choose from the same book?
(b) hen would choose two from each of the books?
(c) hen want at least a hymn from each book?
Note. that the order of the psalms has no significance.

Regards,
\(\displaystyle |\rangle\)

 

[I like Serena]

Hello MHB,
I got difficult understand this kind of questions, anyone got any tips or could explain, cause I find it really hard to understand when I do this kind of problem, this is an old problem from exam that I hopefully did translate well.
A priest needs to choose six hymns for Sunday's show. Hen has three hymnbooks, each with 25
hymns (thus 75 different hymns together). In how many ways can hen choose the hymns of
(a) all the hymns to choose from the same book?
(b) hen would choose two from each of the books?
(c) hen want at least a hymn from each book?
Note. that the order of the psalms has no significance.

Regards,
\(\displaystyle |\rangle\)

Hi $|\pi\rangle$! :)These questions are about combinatorics.
The key concept is the binomial coefficient, written as $\begin{pmatrix}n \\ k\end{pmatrix}$, ${}^nC_k$, nCr, C(n,k), or something similar.

The binomial coefficient $\begin{pmatrix}n \\k\end{pmatrix}$ represents the number of combinations of k items from a collection of n items, assuming the order is not relevant.

For (a) we want to know the number of combinations of 6 psalms from 1 book with 25 psalms.
Since ordering is not relevant, this is $\begin{pmatrix}25 \\ 6\end{pmatrix}$.

 

[Petrus]

Hi $|\pi\rangle$! :)These questions are about combinatorics.
The key concept is the binomial coefficient, written as $\begin{pmatrix}n \\ k\end{pmatrix}$, ${}^nC_k$, nCr, C(n,k), or something similar.

The binomial coefficient $\begin{pmatrix}n \\k\end{pmatrix}$ represents the number of combinations of k items from a collection of n items, assuming the order is not relevant.

For (a) we want to know the number of combinationsof 6 psalms from 1 book with 25 psalms.
Since ordering is not relevant, this is $\begin{pmatrix}25 \\ 6\end{pmatrix}$.

Hello I like Serena :D
I did also did think it was so but the facit says for a) \(\displaystyle 3\begin{pmatrix}25 \\ 6\end{pmatrix}\), I don't understand where 3 comes from

Regards,

$|\pi\rangle$

 

[I like Serena]

Hello I like Serena :D
I did also did think it was so but the facit says for a) \(\displaystyle 3\begin{pmatrix}25 \\ 6\end{pmatrix}\), I don't understand where 3 comes from

Regards,

$|\pi\rangle$

Hen can choose the psalms in \(\displaystyle \begin{pmatrix}25 \\ 6\end{pmatrix}\) ways from the 1st book.
He has the same number of choices from the second and third book.

 

[Petrus]

How would b) works? What I think is that \(\displaystyle \begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\) I don't know if I am thinking correct but, we need 2 from 1,2,3 book that means we got 6 psalms that he need?Regars,
$|\pi\rangle$

 

[I like Serena]

How would b) works? What I think is that \(\displaystyle \begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\) I don't know if I am thinking correct but, we need 2 from 1,2,3 book that means we got 6 psalms that he need?Regars,
$|\pi\rangle$

Yep! That's it!

 

[Petrus]

Yep! That's it!

I am thinking correct on c? Is it three case?
1 1 4
1 2 3
2 2 2

Regards
\(\displaystyle |\pi\rangle\)

 

[I like Serena]

I am thinking correct on c? Is it three case?
1 1 4
1 2 3
2 2 2

Regards
\(\displaystyle |\pi\rangle\)

Yes...

- Εδώ είναι

 

[Petrus]

Yes...

so for case 1:
1-1-4
\(\displaystyle \begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 4\end{pmatrix}\)
case 2:
1-2-3
\(\displaystyle \begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 3\end{pmatrix}\)

case 3:
2-2-2
\(\displaystyle \begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\)
then I add them all?

Regards,
\(\displaystyle |\pi\rangle\)

 

[I like Serena]

For case 1 you took 4 psalms from the 3rd book.
Could you also take 4 psalms from the 1st book?

 

[Petrus]

For case 1 you took 4 psalms from the 3rd book.
Could you also take 4 psalms from the 1st book?

yeah but then i will be able to take 3 from the other two as well so I got:
\(\displaystyle 3 \begin{pmatrix}25 \\ 1\end{pmatrix}3 \begin{pmatrix}25 \\ 1\end{pmatrix}3 \begin{pmatrix}25 \\ 4\end{pmatrix}\)

Regards,
\(\displaystyle |\pi\rangle\)

 

[I like Serena]

yeah but then i will be able to take 3 from the other two as well so I got:
\(\displaystyle 3 \begin{pmatrix}25 \\ 1\end{pmatrix}3 \begin{pmatrix}25 \\ 1\end{pmatrix}3 \begin{pmatrix}25 \\ 4\end{pmatrix}\)

Regards,
\(\displaystyle |\pi\rangle\)

Not quite.
You can indeed pick 4 from the first book (combination 4-1-1) for $\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 4\end{pmatrix}$ possibilities.
Or 4 from the second book (1-4-1) for another $\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 4\end{pmatrix}$ possibilities.
Or 4 from the third book (1-1-4).

Summing those up, I get a total of $3\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 4\end{pmatrix}$ possibilities for case 1.

 

[Petrus]

Not quite.
You can indeed pick 4 from the first book (combination 4-1-1) for $\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 4\end{pmatrix}$ possibilities.
Or 4 from the second book (1-4-1) for another $\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 4\end{pmatrix}$ possibilities.
Or 4 from the third book (1-1-4).

Summing those up, I get a total of $3\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 4\end{pmatrix}$ possibilities for case 1.

how will case two work?
3xx I can have 3 at beginning in 2 way (I mean 321,312)
x3x I can have 3 at second one 2 way
xx3 I can have 3 at third one 2 way
that means I got \(\displaystyle 6\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 3\end{pmatrix}\)??

Regards,
\(\displaystyle |\pi\rangle\)

 

[I like Serena]

Yep!

 

[Petrus]

Yep!

Thanks I like Serena for taking your time and helping me!:) I start to understand this a lot better!(Cool)

Regards,
\(\displaystyle |\pi\rangle\)