Perms and combs (1 Viewer)

[deswa1]

Hey guys,

Could anyone help solve this perms and combs question? - my answer is out by a factor of two and I don't know why:

Nine players are to be divided into two teams of four and one umpire.
a) In how many ways can the teams be formed?
b) If two particular people cannot be on the same team, how many different combinations are possible?

Solved- thanks. There's a new one at the bottom...

NEW QUESTION (also in its seperate post):
The six faces of a number of identical cubes are painted in six distinct colours. How many cubes can be formed?

Thanks heaps

As I find more questions that I can't do- I'll just add to this thread

 

[Sy123]

What does the answer say?

 

[jackerino]

Hey guys,

Could anyone help solve this perms and combs question? - my answer is out by a factor of two and I don't know why:

Nine players are to be divided into two teams of four and one umpire.
a) In how many ways can the teams be formed?
b) If two particular people cannot be on the same team, how many different combinations are possible?

Thanks heaps

As I find more questions that I can't do- I'll just add to this thread

a) 9C4+9C4+soething I think lol.
b) 8C4.9C4. something


sorry not sure

 

[deswa1]

What does the answer say?

315 and 210

 

[Sy123]

Is it you getting 630 the problem for the first one?

 

[deswa1]

Yep. I did (9C4)(5C4)

 

[jackerino]

wait wtf isnt it 9C4.5C1.1C1

 

[Aesytic]

for the first one, the answer is 9C4*5C4*1C1 divided by 2
i assume you know why you multiply those combinations, but you divide by 2 because the teams aren't distinguishable from one another

for the second one, we can assume 3 conditions
let the 2 players be A and B
first condition: A is an umpire
in this case, it's 1C1*8C4*4C4 divided by 2, which is 35, since once again the teams are indistinguishable

second condition: B is an umpire
the result we get is the same as the first condition, so it's 35

third condition: A and B are on different teams
in this case the teams are distinguishable with A on one team and B on the other
because we already have one player in each team, for this calculation it's 7C3*4C3*1C1 = 140

adding these three conditions together, we get 210 combinations

 

[jackerino]

Oh yea its 9C4x5C1x5C4

 

[jackerino]

No demmit got it wrong haha. Well yea aesytic has it.

 

[deswa1]

for the first one, the answer is 9C4*5C4*1C1 divided by 2
i assume you know why you multiply those combinations, but you divide by 2 because the teams aren't distinguishable from one another

for the second one, we can assume 3 conditions
let the 2 players be A and B
first condition: A is an umpire
in this case, it's 1C1*8C4*4C4 divided by 2, which is 35, since once again the teams are indistinguishable

second condition: B is an umpire
the result we get is the same as the first condition, so it's 35

third condition: A and B are on different teams
in this case the teams are distinguishable with A on one team and B on the other
because we already have one player in each team, for this calculation it's 7C3*4C3*1C1 = 140

adding these three conditions together, we get 210 combinations

Oh cool thanks heaps

 

[bleakarcher]

for the first one, the answer is 9C4*5C4*1C1 divided by 2
i assume you know why you multiply those combinations, but you divide by 2 because the teams aren't distinguishable from one another

for the second one, we can assume 3 conditions
let the 2 players be A and B
first condition: A is an umpire
in this case, it's 1C1*8C4*4C4 divided by 2, which is 35, since once again the teams are indistinguishable

second condition: B is an umpire
the result we get is the same as the first condition, so it's 35

third condition: A and B are on different teams
in this case the teams are distinguishable with A on one team and B on the other
because we already have one player in each team, for this calculation it's 7C3*4C3*1C1 = 140

adding these three conditions together, we get 210 combinations

wait yeah this guy is right.

 

[deswa1]

Can I ask another one:

By considering its prime factorisation, find the number of positive divisors of 315,000. To make it easier- its prime factorisation is (2^3)(3^2)(5^4)(7)

 

[iBibah]

lol asked in 2006: http://community.boredofstudies.org/showthread.php?t=108658

What is the answer?

 

[deswa1]

lol asked in 2006: http://community.boredofstudies.org/showthread.php?t=108658

What is the answer?

120. That one just gives the prime factorisation haha- I can get that I just don't know where to go from there...

 

[jackerino]

Woaah dude watch out, seanieg89 and Drongonski are here, 'dey r da maffs gurus, u is gonna have no problem nao LOL

 

[bleakarcher]

Can I ask another one:

By considering its prime factorisation, find the number of positive divisors of 315,000. To make it easier- its prime factorisation is (2^3)(3^2)(5^4)(7)

Say you want to consider all four bases (i.e. 2,3,5,7), the number of factors are 3*2*4*1=24
Say you to consider a particular 3-member combination of these four, the number of factors will be 3*2*4+2*4*1+3*4*1+3*2*1=50
Say you want to consider a particular 2-member combination of these four, the number of factors will be 3*2+3*4+3*1+2*4+2*1+4*1=35
Say you want to consider a particular 1-member combination of these four, the number of factors will be 3+2+4+1=11
so you can have 120 different factors.

 

[deswa1]

@bleak- THANK YOU

That makes perfect sense- lol need to practice a lot on these things

 

[bleakarcher]

@bleak- THANK YOU

That makes perfect sense- lol need to practice a lot on these things

no worries man

 

[deswa1]

Latest question (sorry guys...):

The six faces of a number of identical cubes are painted in six distinct colours. How many cubes can be formed?

Thanks again