Permutation question
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how you get the 25? the 630 is correct...Originally posted by Diehard_89:hmm thanks guys
I don't have the final answer but I get different answer
what I do was
I find the one which has no restriction = 630
and I find the no. of ways whereby the red and blacks are together, meaning at one point one of the reds is in contact with the black = 25
so in the end I get 605 -
pls explain your method....seems like TS have no answer so we have to convince each other thru our methods....Originally posted by c0mplex:answer should be 70... i'm 90% sure... -
that's what i said.Originally posted by Diehard_89:hmm thanks guys
I don't have the final answer but I get different answer
what I do was
I find the one which has no restriction = 630
and I find the no. of ways whereby the red and blacks are together, meaning at one point one of the reds is in contact with the black = 25
so in the end I get 605 -
hmmmOriginally posted by c0mplex:how you get the 25? the 630 is correct...
I group the 4 red and 1 black together and so 5! within the group
then with this group and the remaining 4 white I get 5!
so I get (5!)(5!)/(4!)(4!) = 25 -
really? but I am not sure whether 25 is correctOriginally posted by peeehead:that's what i said.
maybe wrong also
haii
anyway really thanks guys -
total number of ways = 9!/(4!4!) = 630Originally posted by hegu:pls explain your method....seems like TS have no answer so we have to convince each other thru our methods....
treating RB as 1 'ball', number of ways = 8!/(4!3!)x2 = 560
required no. of ways = 630 - 560 = 70
your method is logical too... hmm, different answer... -
by common sense there ought to be more than 25.Originally posted by Diehard_89:hmm thanks guys
I don't have the final answer but I get different answer
what I do was
I find the one which has no restriction = 630
and I find the no. of ways whereby the red and blacks are together, meaning at one point one of the reds is in contact with the black = 25
so in the end I get 605
becos:
no. of ways where red and black together = no. of ways where white and black together (since red and white have the same number) = x (lets say)
x + x must definitely be more than 630 since it covers all the cases plus some overlapping combinations
25 + 25 = 50?
where is black ball in the rest of the (less than) 580 combis? next to green ball ah
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you are assuming that the black balls only resides within the 4 red balls, which may not be the case.Originally posted by Diehard_89:hmmm
I group the 4 red and 1 black together and so 5! within the group
then with this group and the remaining 4 white I get 5!
so I get (5!)(5!)/(4!)(4!) = 25
in other words you missed out cases like:
B R W W W W R R R -
Originally posted by c0mplex:
you cant treat RB as 1 'ball' and permute, this will result in lots of overlap. (that's why your answer of 560 is so large)
total number of ways = 9!/(4!4!) = 630
treating RB as 1 'ball', number of ways = 8!/(4!3!)x2 = 560
required no. of ways = 630 - 560 = 70
your method is logical too... hmm, different answer...
lets see why:
one of your perms: W (R B) R W W R R (where R B is a group)
another of your perms: W R (B R) W W R R (where B R is a group)
err...arent these 2 ways the same?
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ohh ya... i forgot there are sooo many black balls... hahahaha... then you should be correct...175... apology for wrong answer...
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ok I think this should be correctOriginally posted by hegu:case 1:
(W B W) (6 other balls) => 7 groups to permutate altogether. Treat (W B W) as one group
combi = 7!/(2! x 4!) = 105 (need to divide becos same coloured balls are identical)
case 2a:
(B W) (7 other balls) => 7 groups to permutate altogether. (B W) group is fixed on the LHS
case 2a:
(7 other balls) (W B) => 7 groups to permutate altogether. (W B) group is fixed on the RHS
summing up case 2a and 2b, combi = 7!/(3! x 4!) x 2 = 70
add these 3 cases you will get 175.
correct me if i am wrong....or i missed out any cases or overlaps
but what if got a lot of whites? meaning we have to do one by one for each white? -
hmm... so find the number of overlaps... for the BR part, consider RBR now... so number of ways = 105... so that's the missing 105!Originally posted by hegu:you cant treat RB as 1 'ball' and permute, this will result in lots of overlap. (that's why your answer of 560 is so large)
lets see why:
one of your perms: W (R B) R W W R R (where R B is a group)
another of your perms: W R (B R) W W R R (where B R is a group)
err...arent these 2 ways the same?
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you need not do this... all the white balls are identical...Originally posted by Diehard_89:ok I think this should be correct
but what if got a lot of whites? meaning we have to do one by one for each white? -
haha...i guess it's case closed then...Originally posted by c0mplex:hmm... so find the number of overlaps... for the BR part, consider RBR now... so number of ways = 105... so that's the missing 105!
phew what a tough qn ...
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sorry another question but this is combination
2 parallel lines
one line got 2 points P and Q
the other got 8 points
how many triangles could be formed using 3 of the 10 points as vertices
not sure whether my answer is correct so I am asking -
oh mann where is hegu??
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Haven't touched maths in 2 years (darn army, haha.) Could be...
(2C1 x 8C2 ) + (2C2 x 8C1)
??? Not very sure, honestly. And no calculator by my side. -
haha yeapOriginally posted by JerzZzzZ:Haven't touched maths in 2 years (darn army, haha.) Could be...
(2C1 x 8C2 ) + (2C2 x 8C1)
??? Not very sure, honestly. And no calculator by my side.
my question solved
you are right!
thanks! -
No problem, glad I still have abit of maths in me. Now I'll be crushed if someone comes along and says I did it totally wrong.
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Originally posted by JerzZzzZ:No problem, glad I still have abit of maths in me. Now I'll be crushed if someone comes along and says I did it totally wrong.
army got so cham? -
Not cham, but when you stop studying for 2 years, you really forget a lot of things you once mugged, hehe.
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haha
another question
how many ways can 5 copies of a book be distrubited among 10 people? -
Library loan, email ( electronic copy) , snail mail , courier delivered, personal delivery etc.Originally posted by Diehard_89:haha
another question
how many ways can 5 copies of a book be distrubited among 10 people? -
err how can 5 copies of a book be distributed among 10 people?Originally posted by Diehard_89:haha
another question
how many ways can 5 copies of a book be distrubited among 10 people?
unless they tear the book into half?
or 10P5?