MHB How many different research teams can the director build?

[evinda]

Hello!

I am given the following exercise:

The director of a research institute has in possession $6$ mathematicians and $7$ physicists.A research program requires the participation of $3$ mathematicians and $4$ physicists.How many different research teams can the director build,given that $2$ specific physicists deny to cooperate?

I thought that it could be like that: $\binom{6}{3} \cdot \binom{7-2}{4}$
Is it right or am I wrong?

 

[Plato2]

The director of a research institute has in possession $6$ mathematicians and $7$ physicists.A research program requires the participation of $3$ mathematicians and $4$ physicists.How many different research teams can the director build,given that $2$ specific physicists deny to cooperate?
I thought that it could be like that: $\binom{6}{3} \cdot \binom{7-2}{4}$
Is it right or am I wrong?

With no restrictions at all there are $\dbinom{6}{3}\dbinom{7}{4}$ ways to form the group.

Now that are $\dbinom{6}{3}\dbinom{5}{2}$ ways to form the group with the two non-cooperative physicists together on it. But that is exactly what we must avoid.

How do we use those two to answer the question?

 

[I like Serena]

Hello!

I am given the following exercise:

The director of a research institute has in possession $6$ mathematicians and $7$ physicists.A research program requires the participation of $3$ mathematicians and $4$ physicists.How many different research teams can the director build,given that $2$ specific physicists deny to cooperate?

I thought that it could be like that: $\binom{6}{3} \cdot \binom{7-2}{4}$
Is it right or am I wrong?

Hey! :)

It seems you don't want either of those 2 physicists on your team! (Wait)
But I think it would be okay if you just have 1 of those physicists in the team...

 

[evinda]

Hey! :)

It seems you don't want either of those 2 physicists on your team! (Wait)
But I think it would be okay if you just have 1 of those physicists in the team...

Oh yes,right! (Nod) But how can I calculate then the number of physicists that could be in a team??

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Now that are $\dbinom{6}{3}\dbinom{5}{2}$ ways to form the group with the two non-cooperative physicists together on it. But that is exactly what we must avoid.

Could you explain me further what this $\dbinom{6}{3}\dbinom{5}{2}$ stands for? (Wondering)

 

[I like Serena]

Oh yes,right! (Nod) But how can I calculate then the number of physicists that could be in a team??

Let's start with the number of physicist teams that won't work out if those 2 physicists, that are so difficult, would be in it.
Can you calculate that?

 

[Plato2]

Oh yes,right! (Nod) But how can I calculate then the number of physicists that could be in a team??

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Could you explain me further what this $\dbinom{6}{3}\dbinom{5}{2}$ stands for?

$\dbinom{6}{3}$ is the number of ways to choose the mathematicians.

$\dbinom{5}{2}$ is the number of ways to choose the two of the of the five cooperating physicists to put with the two who will not cooperate.

In other words $\dbinom{6}{3}\dbinom{5}{2}$ is the number of team we do not want.

Subtract that from the total number of possible teams.

 

[evinda]

Let's start with the number of physicist teams that won't work out if those 2 physicists, that are so difficult, would be in it.
Can you calculate that?

As two physicists don't want to cooperate, we don't want $\binom{5}{2}$ physicists,or don't we take into consideration then that two specific don't want to cooperate??

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$\dbinom{6}{3}$ is the number of ways to choose the mathematicians.

$\dbinom{5}{2}$ is the number of ways to choose the two of the of the five cooperating physicists to put with the two who will not cooperate.

In other words $\dbinom{6}{3}\dbinom{5}{2}$ is the number of team we do not want.

Subtract that from the total number of possible teams.

So, $\dbinom{5}{2}$ expresses the two specific physicists,that don't want to cooperate,or just any two ones?

 

[I like Serena]

As two physicists don't want to cooperate, we don't want $\binom{5}{2}$ physicists,or don't we take into consideration then that two specific don't want to cooperate??

Let's take into consideration that 2 specific ones, say Anna and Berta, do not want to cooperate.

 

[evinda]

Let's take into consideration that 2 specific ones, say Anna and Berta, do not want to cooperate.

I don't really know...So, $\binom{7}{4}-\binom{4}{2}$ physicists are possible to me in a team??

 

[Plato2]

I don't really know...So, $\binom{7}{4}-\binom{4}{2}$ physicists are possible to me in a team??

O.K. Admittedly we are making several assumptions about the setup (the wording) of your question.
Both of us assume that there are two physicists who will not serve together, call them A & B.
That means that any research group cannot have both A & B among the seven.
Now a group can have A but not B, B but not A, or not A and not B.

The number of ways that a research group can contain A and not B is $\dbinom{6}{3}\dbinom{5}{3}.$

The number of ways that a research group can contain B and not A is $\dbinom{6}{3}\dbinom{5}{3}.$

The number of ways that a research group can contain not A and not B is $\dbinom{6}{3}\dbinom{5}{4}.$

If you add those three numbers together you will see it is $\dbinom{6}{3}\dbinom{7}{4}-\dbinom{6}{3}\dbinom{5}{2}$

 

[I like Serena]

So, $\dbinom{5}{2}$ expresses the two specific physicists,that don't want to cooperate,or just any two ones?

The number $\dbinom{7-2}{2} = \dbinom{5}{2}$ expresses any 2 physicists that are not the 2 specific physicists.
That is, we pick 2 physicists from the remaining 5 and we pair them up with A and B, so we have a team of 4 physicists.

 

[evinda]

The number $\dbinom{7-2}{2} = \dbinom{5}{2}$ expresses any 2 physicists that are not the 2 specific physicists.
That is, we pick 2 physicists from the remaining 5 and we pair them up with A and B, so we have a team of 4 physicists.

The number $\dbinom{7-2}{2} = \dbinom{5}{2}$ expresses any 2 physicists that are not the 2 specific physicists.
That is, we pick 2 physicists from the remaining 5 and we pair them up with A and B, so we have a team of 4 physicists.

If we consider that the physicists are these:

$P1,P2,P3,P4,P5,P6,P7$

and $P6,P7$ don't want to cooperate, aren't the couples, except from $P6 \text{ with} P7$ these?

• P1-P2
• P1-P3
• P1-P4
• P1-P5
• P1-P6
• P1-P7
• P2-P3
• P2-P3
• P2-P4
• P2-P5
• P2-P6
• P2-P7
• P3-P4
• P3-P5
• P3-P6
• P3-P7
• P4-P5
• P4-P6
• P4-P7
• P5-P6
• P5-P7

So,there are $20$ possible couples...But, $ \dbinom{5}{2}=\frac{5!}{2!3!}=10$..Doesn't $ \dbinom{5}{2}$ stand for the possible couples,except from the two,that do not want to cooperate? Or have I understood it wrong?? (Blush)

 

[I like Serena]

We intended the number of teams that won't work out:

1. P1-P2-P6-P7
2. P1-P3-P6-P7
3. P1-P4-P6-P7
4. P1-P5-P6-P7
5. P2-P3-P6-P7
6. P2-P4-P6-P7
7. P2-P5-P6-P7
8. P3-P4-P6-P7
9. P3-P5-P6-P7
10. P4-P5-P6-P7

As you can see these are $\dbinom{5}{2}=10$ possibilities that always include $P6-P7$. (Bandit)

Since the total number of physicist teams is $\dbinom{7}{4}$, the number of allowed teams is:
$$\dbinom{7}{4} - \dbinom{5}{2} = 35 - 10 = 25$$
(Sun)

 

[evinda]

We intended the number of teams that won't work out:

1. P1-P2-P6-P7
2. P1-P3-P6-P7
3. P1-P4-P6-P7
4. P1-P5-P6-P7
5. P2-P3-P6-P7
6. P2-P4-P6-P7
7. P2-P5-P6-P7
8. P3-P4-P6-P7
9. P3-P5-P6-P7
10. P4-P5-P6-P7

As you can see these are $\dbinom{5}{2}=10$ possibilities that always include $P6-P7$. (Bandit)

Since the total number of physicist teams is $\dbinom{7}{4}$, the number of allowed teams is:
$$\dbinom{7}{4} - \dbinom{5}{2} = 35 - 10 = 25$$
(Sun)

I think I understood it! So,the director can build $25 \cdot 20=500$ research teams,right?

 

[I like Serena]

I think I understood it! So,the director can build $25 \cdot 20=500$ research teams,right?

Yep! (Mmm)

 

[evinda]

O.K. Admittedly we are making several assumptions about the setup (the wording) of your question.
Both of us assume that there are two physicists who will not serve together, call them A & B.
That means that any research group cannot have both A & B among the seven.
Now a group can have A but not B, B but not A, or not A and not B.

The number of ways that a research group can contain A and not B is $\dbinom{6}{3}\dbinom{5}{3}.$

The number of ways that a research group can contain B and not A is $\dbinom{6}{3}\dbinom{5}{3}.$

The number of ways that a research group can contain not A and not B is $\dbinom{6}{3}\dbinom{5}{4}.$

If you add those three numbers together you will see it is $\dbinom{6}{3}\dbinom{7}{4}-\dbinom{6}{3}\dbinom{5}{2}$

Yep! (Mmm)

Nice,thank you!

 

[Fermat1]

I like to think of the 2 physicists as a separate group. The options are taking neither of these 2, or taking one. So the number of possible teams is $6C3.5C4+6C3.5C3.2C1$, the first term corresponding to taking neither of the two refusers, the second term corresponding to taking (exactly) 1.