# Combinatorics

#### Ninjapancake37

##### New member
Nor sure if combinatorics counts as advanced math or not, but... I am currently at a summer camp and we have 10 award requirements on a card to fulfill each requiring 1 signature. Their are a total of 14 people authorized to sign off on a maximum of 2 requirements each, so here is my question: how many combinations are there total of complete and incomplete requirement cards?

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#### Ninjapancake37

##### New member
Nor sure if combinatorics counts as advanced math or not, but... I am currently at a summer camp and we have 10 award requirements on a card to fulfill each requiring 1 signature. Their are a total of 14 people authorized to sign off on a maximum of 2 requirements each, so here is my question: how many combinations are there total of complete and incomplete requirement cards?
Also all I really now about this is that it relates to combinatorics. I only no about it because I watch numberphile on YouTube and it isn’t my homework; I’m just doing this for fun.

#### pka

##### Elite Member
10 award requirements on a card to fulfill each requiring 1 signature. Their are a total of 14 people authorized to sign off on a maximum of 2 requirements each, so here is my question: how many combinations are there total of complete and incomplete requirement cards?
I delayed trying to compose an answer hoping Prof. Peterson might give us his reading. He is better at that than I. Here is what I understand you to mean. Each judge can sign a card a maximum of two times. Hence a complete card can have a minimum of five unique signatures and a maximum of ten. Thus cards run from being empty to having all events filled. Please have a look at this calculation. In that expanded polynomial there is a $$\displaystyle 1$$ telling us the is only one way to have an empty card, no signatures. The $$\displaystyle 14x$$ means there are fourteen ways to have a card with one signature. The $$\displaystyle 105x^2$$ says that there $$\displaystyle 91+14=105$$ ways to have a card with two events completed: that is $$\displaystyle \binom{14}{2}=91$$ fourteen choose two +fourteen. So $$\displaystyle \bf 270270x^{10}$$ is gives the number of different complete cards.

#### Ninjapancake37

##### New member
I delayed trying to compose an answer hoping Prof. Peterson might give us his reading. He is better at that than I. Here is what I understand you to mean. Each judge can sign a card a maximum of two times. Hence a complete card can have a minimum of five unique signatures and a maximum of ten. Thus cards run from being empty to having all events filled. Please have a look at this calculation. In that expanded polynomial there is a $$\displaystyle 1$$ telling us the is only one way to have an empty card, no signatures. The $$\displaystyle 14x$$ means there are fourteen ways to have a card with one signature. The $$\displaystyle 105x^2$$ says that there $$\displaystyle 91+14=105$$ ways to have a card with two events completed: that is $$\displaystyle \binom{14}{2}=91$$ fourteen choose two +fourteen. So $$\displaystyle \bf 270270x^{10}$$ is gives the number of different complete cards.

Wouldn’t there be 14*10 ways to have one signature sense there are 14 different people each with 1 of 10 spaces they could possibly write their signature? They don’t have to sign in any particular order.

#### Dr.Peterson

##### Elite Member
The first thing we need is to be sure what we are counting.

Let's restate the problem. We could say that we want to fill in 10 blanks, _ _ _ _ _ _ _ _ _ _, with letters from A to N (14 possibilities), allowing no more than two of any given letter. There must be a letter in each space, and it matters where each letter is. How many such ten-letter "words" are there?

Does that sound equivalent to your problem?

Combinatorics is a field in which it is very easy to make up, state, and even understand a very hard problem! My guess is that you will need to use generating functions, which I know very little about (but pka knows well, so I'll leave that to him).

#### pka

##### Elite Member
we want to fill in 10 blanks, _ _ _ _ _ _ _ _ _ _, with letters from A to N (14 possibilities), allowing no more than two of any given letter. There must be a letter in each space, and it matters where each letter is. How many such ten-letter "words" are there?
According to the OP, the ten blanks can have anywhere from zero to ten blanks filled, with no more than two of the fourteen possible. I think that the OP wants to ask about content (i.e. not about who signs off in what but how many possible configurations are there?) Each adjudicator presented a card can choose to sign not sign for any even, to sign for one event, or for two of the ten events.

#### Dr.Peterson

##### Elite Member
I am currently at a summer camp and we have 10 award requirements on a card to fulfill each requiring 1 signature. Their are a total of 14 people authorized to sign off on a maximum of 2 requirements each, so here is my question: how many combinations are there total of complete and incomplete requirement cards?
According to the OP, the ten blanks can have anywhere from zero to ten blanks filled, with no more than two of the fourteen possible. I think that the OP wants to ask about content (i.e. not about who signs off in what but how many possible configurations are there?) Each adjudicator presented a card can choose to sign not sign for any even, to sign for one event, or for two of the ten events.
This is why I asked the OP for confirmation or correction. I can't be sure.

Looking now, I see "complete and incomplete", which I must have missed originally. That makes it even more complicated. I was considering only full cards. You are right on that aspect; it's not just 10-letter words, unless perhaps you allow as many X's (unsigned) as necessary.

But I do think it matters who signed off on which requirements, as that would make a difference in the cards. I don't take the word "combinations" literally, though it could be; I take the requirements as being distinct.

#### Ninjapancake37

##### New member
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This is why I asked the OP for confirmation or correction. I can't be sure.

Looking now, I see "complete and incomplete", which I must have missed originally. That makes it even more complicated. I was considering only full cards. You are right on that aspect; it's not just 10-letter words, unless perhaps you allow as many X's (unsigned) as necessary.

But I do think it matters who signed off on which requirements, as that would make a difference in the cards. I don't take the word "combinations" literally, though it could be; I take the requirements as being distinct.
That‘s correct I was referring to both complete and incomplete cards. Also pka you are right about the content of combinations.

#### Ninjapancake37

##### New member
This is why I asked the OP for confirmation or correction. I can't be sure.

Looking now, I see "complete and incomplete", which I must have missed originally. That makes it even more complicated. I was considering only full cards. You are right on that aspect; it's not just 10-letter words, unless perhaps you allow as many X's (unsigned) as necessary.

But I do think it matters who signed off on which requirements, as that would make a difference in the cards. I don't take the word "combinations" literally, though it could be; I take the requirements as being distinct.
I think the solution is to add a fifteenth letter “O” representing a blank. And if that is the case given everything else what formula or equation would I use to solve this problem? Could you give me an example? Also I am not familiar with the parentheses around the two numbers on bottom and top. Could you fill me in?

#### pka

##### Elite Member
I am not familiar with the parentheses around the two numbers on bottom and top. Could you fill me in?
From twenty choose six is $$\displaystyle \dbinom{20}{6}=\dfrac{20!}{6!(20-6)!}$$
From N choose k is $$\displaystyle \dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}$$
Bookmark this site and use it.

#### Dr.Peterson

##### Elite Member
I think the solution is to add a fifteenth letter “O” representing a blank. And if that is the case given everything else what formula or equation would I use to solve this problem? Could you give me an example? Also I am not familiar with the parentheses around the two numbers on bottom and top. Could you fill me in?
Yes, your O is what I suggested as X to make it stand out more. So we are looking for all 10-letter words consisting of no more than 2 each of letters A through N, together with any number (0 to 10) of X. Is that right?

I think, again, that pka will be the one to provide the method, which will probably be generating functions, which he used in post #3 without explanation. I'm not sufficiently knowledgeable about the subject to give a confident answer.

And, as he said, $$\displaystyle \binom{n}{r}$$ is a notation for combinations ("n choose r") commonly used where you may be more familiar with something like $$\displaystyle _nC_r$$.