Combinatorics Word Problem: Choosing Hymns for a Sunday Service

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Discussion Overview

The discussion revolves around a combinatorial problem involving the selection of hymns for a Sunday service. Participants explore different methods to calculate the number of ways to choose hymns from multiple hymnbooks, focusing on various conditions such as selecting all hymns from the same book, choosing a specific number from each book, and ensuring at least one hymn is selected from each book.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants discuss the use of the binomial coefficient to determine combinations, specifically noting that for selecting 6 hymns from one book with 25 hymns, the calculation is represented as $\begin{pmatrix}25 \\ 6\end{pmatrix}$.
  • Others propose that the total number of ways to choose all hymns from the same book should be multiplied by 3, leading to the expression $3\begin{pmatrix}25 \\ 6\end{pmatrix}$, although the reasoning behind the factor of 3 is questioned.
  • For the second part of the problem, participants suggest that the number of ways to choose 2 hymns from each of the three books can be expressed as $\begin{pmatrix}25 \\ 2\end{pmatrix}^3$, indicating agreement on this approach.
  • In discussing the third part, participants identify three cases for selecting hymns: (1) 1 from each of the first two books and 4 from the third, (2) 1 from one book, 2 from another, and 3 from the last, and (3) 2 from each book.
  • Some participants express uncertainty about how to calculate the total combinations for each case, leading to further exploration of the combinations involved.
  • There is a discussion about the possibility of selecting 4 hymns from different books and how to account for the different arrangements of selections across the books.
  • Participants also explore the permutations of selections in the second case, leading to a proposed total of $6\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 3\end{pmatrix}$.

Areas of Agreement / Disagreement

Participants generally agree on the methods for calculating combinations using binomial coefficients, but there is some disagreement regarding the application of these methods to specific cases and the reasoning behind certain multipliers. The discussion remains unresolved in terms of final calculations and interpretations.

Contextual Notes

Some assumptions regarding the selection process and the interpretation of the problem may not be fully articulated, leading to varying approaches in the calculations. The discussion also reflects a dependence on the definitions of combinations and the specific conditions outlined in the problem.

Who May Find This Useful

This discussion may be useful for individuals interested in combinatorial mathematics, particularly in understanding how to approach problems involving selections from multiple groups under specific constraints.

Petrus
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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,
$$|\rangle$$
 
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Petrus said:
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,
$$|\rangle$$

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 $\begin{pmatrix}25 \\ 6\end{pmatrix}$.
 
Last edited:
I like Serena said:
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) $$3\begin{pmatrix}25 \\ 6\end{pmatrix}$$, I don't understand where 3 comes from

Regards,

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

Regards,

$|\pi\rangle$

Hen can choose the psalms in $$\begin{pmatrix}25 \\ 6\end{pmatrix}$$ ways from the 1st book.
He has the same number of choices from the second and third book.
 
How would b) works? What I think is that $$\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}$$ 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,
$|\pi\rangle$
 
Petrus said:
How would b) works? What I think is that $$\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}$$ 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!
 
I like Serena said:
Yep! That's it!
I am thinking correct on c? Is it three case?
1 1 4
1 2 3
2 2 2

Regards
$$|\pi\rangle$$
 
Petrus said:
I am thinking correct on c? Is it three case?
1 1 4
1 2 3
2 2 2

Regards
$$|\pi\rangle$$

Yes...

- Εδώ είναι
 
I like Serena said:
Yes...
so for case 1:
1-1-4
$$\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 4\end{pmatrix}$$
case 2:
1-2-3
$$\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 3\end{pmatrix}$$

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

Regards,
$$|\pi\rangle$$
 
  • #10
For case 1 you took 4 psalms from the 3rd book.
Could you also take 4 psalms from the 1st book?
 
  • #11
I like Serena said:
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:
$$3 \begin{pmatrix}25 \\ 1\end{pmatrix}3 \begin{pmatrix}25 \\ 1\end{pmatrix}3 \begin{pmatrix}25 \\ 4\end{pmatrix}$$

Regards,
$$|\pi\rangle$$
 
  • #12
Petrus said:
yeah but then i will be able to take 3 from the other two as well so I got:
$$3 \begin{pmatrix}25 \\ 1\end{pmatrix}3 \begin{pmatrix}25 \\ 1\end{pmatrix}3 \begin{pmatrix}25 \\ 4\end{pmatrix}$$

Regards,
$$|\pi\rangle$$

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.
 
  • #13
I like Serena said:
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 $$ 6\begin{pmatrix}25 \\ 1\end{pmatrix}\begin{pmatrix}25 \\ 2\end{pmatrix}\begin{pmatrix}25 \\ 3\end{pmatrix}$$??

Regards,
$$|\pi\rangle$$
 
  • #14
Yep!
 
  • #15
I like Serena said:
Yep!
Thanks I like Serena for taking your time and helping me!:) I start to understand this a lot better!(Cool)

Regards,
$$|\pi\rangle$$
 

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