perms and combs (1 Viewer)

[jnney]

a group of 3 men and 7 boys include a father and son. from this group, a team consisting of 2 men and 2 boys are to be chosen. how many ways can the team be chosen if the father and son cannot be on the team together?

 

[Shadowdude]

This may or may not be right:

Case 1: Father on the team

1. Pick the father... 1 way
2. Pick one other 'man'... 2 ways
3. Pick two boys not including the son... C(6,2)

Case 2: Son on the team

1. Pick the two non-father men ... 1 way
2. Pick the son ... 1 way
3. Pick one other boy... 6 ways

Case 3: Neither on the team

1. Pick the two non-father men ... 1 way
2. Pick two boys from the non-son people ... C(6,2)

Answer: 3C(6,2) + 6

---

Right? Right? If not... well. Let it be an exercise to see what I did wrong. Also: post count +1

 

[jnney]

The answer is 51!

 

[someth1ng]

a group of 3 men and 7 boys include a father and son. from this group, a team consisting of 2 men and 2 boys are to be chosen. how many ways can the team be chosen if the father and son cannot be on the team together?

Total Ways=3C2*7C2=63 ways

If Father+Son is not possible then:

Total Ways Satisfying Conditions=3C2*7C2 - Ways Father+Son on the Team
Ways Father+Son on the Team=2C(2-1)*6C(2-1)=12 ways

Total ways=63-12=51 ways

 

[jnney]

Ways Father+Son on the Team=2C(2-1)*6C(2-1)=12

I don't understand this line. Can you please explain it?

 

[someth1ng]

Ways Father+Son on the Team=2C(2-1)*6C(2-1)=12

I don't understand this line. Can you please explain it?

Well, you assume that the case if Father+Son is on the team then there is only 1 man and 1 boy to be picked.
There are 2 men remaining to be picked from and 6 remaining boys to be picked since you have already picked the father/son.

Hence of the 2 men left - 2C1
AND
of the 6 boys left - 6C1

Multiply this value together to get the total ways. ie 6*2=12

Now you know the cases in which the Father+Son are together, you subtract this from the total to find the cases that Father+Son are NOT on the same team.

 

[Shadowdude]

btw, 3C(6,2) + 6 = 51. My solution is nice and valid too <_<

 

[someth1ng]

3C(6,2)

?

 

[Shadowdude]

3C(6,2)?



You should really know the other ways to represent things... C(a,b) is kinda standard notation.

 

[someth1ng]

3C(6,2)?



You should really know the other ways to represent things... C(a,b) is kinda standard notation.

You do realise that notation isn't used in 3U? Writing it as C(6,2) makes it impossible for a 3U student to decipher.

 

[Shadowdude]

Well, what else would it be?

Of course I realise that that notation isn't used in 3u, I did it myself. But 6C3 is less neat, and there's no way of doing the binomial thing in normal text.

 

[someth1ng]

Well, what else would it be?

Of course I realise that that notation isn't used in 3u, I did it myself. But 6C3 is less neat, and there's no way of doing the binomial thing in normal text.

How the hell can you just decipher it without knowing it - obviously. If that notation isn't used in 3U, why used it on OP who just started MX2 and wouldn't have been exposed to such notations and even yet - might not even be exposed to anything that isn't nCr in format.

 

[Shadowdude]

How the hell can you just decipher it without knowing it - obviously. If that notation isn't used in 3U, why used it on OP who just started MX2 and wouldn't have been exposed to such notations and even yet - might not even be exposed to anything that isn't nCr in format.

This is why I set out my working clearly. For example:

"3. Pick two boys not including the son... C(6,2)"

So from 6 people, I am selecting two. So that can be done in "6C2" ways. I think it wouldn't be that much of a mental step to go: "Ohhhh he just made the thing neater by bracketing terms..."

But apparently it is.

 

[asianese]

LOL@ the difference in BOS/HSC maths and real world maths... (conventions and rules and assumptions etc.)

 

[4025808]

Well, what else would it be?

Of course I realise that that notation isn't used in 3u, I did it myself. But 6C3 is less neat, and there's no way of doing the binomial thing in normal text.

How the hell can you just decipher it without knowing it - obviously. If that notation isn't used in 3U, why used it on OP who just started MX2 and wouldn't have been exposed to such notations and even yet - might not even be exposed to anything that isn't nCr in format.

Shadow, in your situation, it's not actually that easy, because the notation is different. Also when many people in uni were exposed to this sort of notation, they went WTF??? 3U and 4U students are not taught this notation at school.

Imo there's not much difference between C(6,3) and 6C3 anyway.

 

[Shadowdude]

What? When we learnt it in Discrete it was: "Oh okay, so that's how you can do it". It wasn't: "WTF IS THIS".

It's logical: nCr -> C(n,r). I don't see the difficulty in it.

I admit that I used nCr notation for about four weeks even after learning C(n,r) but I've found that to be neater and I use it all the time now.