HSC 2015 MX2 Permutations & Combinations Marathon (archive) (1 Viewer)

[Drsoccerball]

Re: 2015 permutation X2 marathon

If im not mistaken, it should be 2/15

The 2! is for the opposiote side

 

[Ekman]

Re: 2015 permutation X2 marathon

The 2! is for the opposiote side

My logic was:

Case 1 (Where the pair are together on the side with 2 seats):
10! is the unrestricted amounts
Put the pair down on the 2 seats, you have 8! for the other people
2! between the pair itself
(And if the question said that each side is different, then 2! again between the two sides with 2 seats)

Case 2 (Where the pair are together on the side with 3 seats):
10! is the unrestricted amounts
Put the pair down on the 2 of the 3 seats, you have 8! for the other people
2! between the pair itself
2! between the pair and the third person
(And if the question said that each side is different, then 2! again between the two sides with 3 seats)

Therefore: 8!*2!(1+2!) /10! = 1/15 (If each side was identical)
And 8! * 2! *2!(1+2!) / 10! = 2/15 (If each side was different)

 

[braintic]

Re: 2015 permutation X2 marathon

The 2! is for the opposiote side

If the 2 people randomly choose seats, the probability of them choosing two adjacent seats on the same side of the table is definitely 1/15 regardless of whether sides are considered identical or different.

I'm not sure what logic you are using to get that answer.

 

[InteGrand]

Re: 2015 permutation X2 marathon

If the 2 people randomly choose seats, the probability of them choosing two adjacent seats on the same side of the table is definitely 1/15 regardless of whether sides are considered identical or different.

I'm not sure what logic you are using to get that answer.

Why isn't it 2/15?

Assume the seats are all different.

There's 10P2 ways to seat the two people.

The number of ways they can be seated next to each other is: 2!•(6), since there's 6 ways to place a pair together on the table, and 2! ways to arrange each person in the pair.

So the probability would be 2!•6/(10P2) = 12/(10•9) = 12/90 = 2/15. ?

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

Why isn't it 2/15?

Assume the seats are all different.

There's 10P2 ways to seat the two people.

The number of ways they can be seated next to each other is: 2!•(6), since there's 6 ways to place a pair together on the table, and 2! ways to arrange each person in the pair.

So the probability would be 2!•6/(10P2) = 12/(10•9) = 12/90 = 2/15. ?

What about the other people ? They're included in the possibility eg: If a an b sit on one side the other sides can be changed making alot more combination

 

[InteGrand]

Re: 2015 permutation X2 marathon

What about the other people ? They're included in the possibility eg: If a an b sit on one side the other sides can be changed making alot more combination

The question didn't mention there are other people. Also, I don't think it makes a difference (for the probability).

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

The question didn't mention there are other people. Also, I don't think it makes a difference.

WOOPS MY BAD D:
My angle of attack was this :
Lets fix them on one of the sides (2 seats )
Combinations : (2C1)2!8!
Fixing them on the other sides :
Combinations 2C1)2!8!x2

 

[mreditor16]

Re: 2015 permutation X2 marathon

New Q -

The Australian selectors are selecting the cricket team for the
next test. They have to chose a team of eleven from 15 possible players. Six
of the players can bowl, three can keep wicket and none can bat, bowl and
keep wicket. The selectors have decided that they need at least 4 bowlers and
at least 2 wicket-keepers in the eleven. How many possible teams do they
have to select from?

 

[InteGrand]

Re: 2015 permutation X2 marathon

New Q -

The Australian selectors are selecting the cricket team for the
next test. They have to chose a team of eleven from 15 possible players. Six
of the players can bowl, three can keep wicket and none can bat, bowl and
keep wicket. The selectors have decided that they need at least 4 bowlers and
at least 2 wicket-keepers in the eleven. How many possible teams do they
have to select from?

Is there anyone who can do exactly two of: bowl, keep wicket. ?

(I'm assuming everyone can bat...lol)

 

[mreditor16]

Re: 2015 permutation X2 marathon

Is there anyone who can do exactly two of: bowl, keep wicket. ?

(I'm assuming everyone can bat...lol)

Nope, because if you're keeping wicket, you don't have a chance to bowl.

 

[InteGrand]

Re: 2015 permutation X2 marathon

Nope, because if you're keeping wicket, you don't have a chance to bowl.

If you're bowling, do you have a chance to keep wicket?

 

[mreditor16]

Re: 2015 permutation X2 marathon

If you're bowling, do you have a chance to keep wicket?

No, besides bowlers do not learn how to keep wicket, and wicker keepers do not learn how to bowl.

 

[InteGrand]

Re: 2015 permutation X2 marathon

No, besides bowlers do not learn how to keep wicket, and wicker keepers do not learn how to bowl.

The point was that the HSC would make it clear what the rules of the game are if a sports-based Q is asked (also since this is a hypothetical perms and combs Q, there could be someone who knows how to do everything unless specified otherwise).

 

[porcupinetree]

Re: 2015 permutation X2 marathon

New Q -

The Australian selectors are selecting the cricket team for the
next test. They have to chose a team of eleven from 15 possible players. Six
of the players can bowl, three can keep wicket and none can bat, bowl and
keep wicket. The selectors have decided that they need at least 4 bowlers and
at least 2 wicket-keepers in the eleven. How many possible teams do they
have to select from?

This is what I'm thinking:

6C4 - choose 4 bowlers from the 6 available
3C2 - choose 2 wicket keepers from the 3 available
9C5 - choose 5 other members from the 9 available

Total: 6C4.3C2.9C5 = 5670

 

[InteGrand]

Re: 2015 permutation X2 marathon

Anyway, we can choose 11 from:

• 3 wicket-keepers (must have at least 2)
• 6 bowlers (must have at least 4)
• 6 others.

If W ∈ {2,3} is the no. of wicket-keepers picked and B ∈ {4,5,6} is the no. of bowlers picked, then we can pick the wicket-keepers in ³C_W ways, the bowlers in ⁶C_B ways, and the remaining 11 – (W + B) people in ⁶C_{11 – (W + B)} ways (note that we always have 0 ≤ 11 – (W + B) ≤ 6, so this binomial coefficient makes sense).

So the no. of ways with W wicket-keepers and B bowlers is

Now we just need to sum this over all 6 possibilities for (W, B), which I'll leave for someone else to do (it's easy, just tedious).

 

[mreditor16]

Re: 2015 permutation X2 marathon

This is what I'm thinking:

6C4 - choose 4 bowlers from the 6 available
3C2 - choose 2 wicket keepers from the 3 available
9C5 - choose 5 other members from the 9 available

Total: 6C4.3C2.9C5 = 5670

That's what I initially fell into the trap of doing as well.

But unfortunately its definitely incorrect, because the answer obtained is more than the number of possibilities without restrictions = 15C11 = 1365.

However, could someone please point out the problem with this working, as I am curious to what makes it wrong....

 

[braintic]

Re: 2015 permutation X2 marathon

Why isn't it 2/15?

Assume the seats are all different.

There's 10P2 ways to seat the two people.

The number of ways they can be seated next to each other is: 2!•(6), since there's 6 ways to place a pair together on the table, and 2! ways to arrange each person in the pair.

So the probability would be 2!•6/(10P2) = 12/(10•9) = 12/90 = 2/15. ?

I had 6/10C2 and somehow thought 10C2 was 90.

Still don't know though why people insist on applying ordering to questions that don't require ordering.

 

[InteGrand]

Re: 2015 permutation X2 marathon

That's what I initially fell into the trap of doing as well.

But unfortunately its definitely incorrect, because the answer obtained is more than the number of possibilities without restrictions = 15C11 = 1365.

However, could someone please point out the problem with this working, as I am curious to what makes it wrong....

This over-counts.

Say the 6 possible bowlers are A,B,C,D,E and F. If we pick 4 bowlers A,B,C and D, then we're allowing ourselves to pick E and F as part of the "others".

Then using this method, we say that this is a DIFFERENT selection: pick C,D,E,F as the "bowlers". Pick A,B as part of the "others".

In the end, both these selection methods can end up choosing the same people, but the selections are incorrectly labelled as "different" because of how we labelled who our "bowlers" are. This is why we need to sum up over cases of picking (4 OR 5 OR 6 bowlers from A,B,...,F) and picking (2 OR 3) wicket-keepers from the list of available wicket-keepers.

 

[InteGrand]

Re: 2015 permutation X2 marathon

I had 6/10C2 and somehow though 10C2 was 90.

Still don't know though why people insist on applying ordering to questions that don't require ordering.

Probably because most people don't find probability/combinatorics intuitive.

 

[mreditor16]

Re: 2015 permutation X2 marathon

Anyway, we can choose 11 from:

• 3 wicket-keepers (must have at least 2)
• 6 bowlers (must have at least 4)
• 6 others.

If W ∈ {2,3} is the no. of wicket-keepers picked and B ∈ {4,5,6} is the no. of bowlers picked, then we can pick the wicket-keepers in ³C_W ways, the bowlers in ⁶C_B ways, and the remaining 11 – (W + B) people in ⁶C_{11 – (W + B)} ways (note that we always have 0 ≤ 11 – (W + B) ≤ 6, so this binomial coefficient makes sense).

So the no. of ways with W wicket-keepers and B bowlers is

Now we just need to sum this over all 6 possibilities for (W, B), which I'll leave for someone else to do (it's easy, just tedious).

Your method gets an answer of 960.