Permutations combinations and probablity

[RahulSharma]

there is if boxes are 2. but beyond that requires some struggle.

[RahulSharma]

say you have 10 red and 20 blue balls and 2 same boxes.
each box has to contain balls of both the colours.

9*19 as usual when boxes are different. (9*19-1)/2 when they are same.

when 10 red, 20 blue, 30 green to be put in 3 same boxes?

(9c2*19c2*29c2 - some number)/3!

same balls same boxes. balls=10, boxes=3

floor(n-1/2) + floor(n-2/2)+.... where n=number of balls

[RahulSharma]

sorry. (9*19 +1)/2

and + some number not -.

[muthuisfine]

hi

for the case(2 2 0 0 0 ) i mean you have put 4 different ball in 5 boxes
and here there are 2 groups so wont the number of ways for this be 4!/(2!*2!)
please correct me if i am wrong

and for the case where u have similar ball to be put in different boxes we can use the formulae (n+r-1)Cr right and in this case n is the number of boxes and r is the number of balls, but whereas u have used (n+r-1)C(r-1) where i think you have taken the number of balls to be n and the number of boxes to be r.
which formulae is right?
the first one or the second one...
please dont get confused..
lest take an example of 7 similar ball to be put in 3 different boxes
as per the formulae (n+r-1)Cr we have r=7 and n=3
so we get 9C7 and whereas your formulae that is(n+r-1)C(r-1) we have r=3 and n=7 we get 9C2.and both have the same value as 9C7 =9C2
i saw the formulae (n+r-1)Cr for this case in an IIT maths text book
Are both the formule the one and the same....do they give the same answer whn tried for different combinations..

also tell if there is a direct formulae or something for the case when different balls
have to be put in similar boxes other than the method you have done..

also tell me when 7 different balls are put in to 3 different boxes
is the number of ways 3^7 or
is it 9!/(2!)

thanks in advance

[Jyotiraj]

[highlight=cadetblue:067b98f9b2]hi

for the case(2 2 0 0 0 ) i mean you have put 4 different ball in 5 boxes
and here there are 2 groups so wont the number of ways for this be 4!/(2!*2!)
please correct me if i am wrong

and for the case where u have similar ball to be put in different boxes we can use the formulae (n+r-1)Cr right and in this case n is the number of boxes and r is the number of balls, but whereas u have used (n+r-1)C(r-1) where i think you have taken the number of balls to be n and the number of boxes to be r.
which formulae is right?
the first one or the second one...
please dont get confused..
lest take an example of 7 similar ball to be put in 3 different boxes
as per the formulae (n+r-1)Cr we have r=7 and n=3
so we get 9C7 and whereas your formulae that is(n+r-1)C(r-1) we have r=3 and n=7 we get 9C2.and both have the same value as 9C7 =9C2
i saw the formulae (n+r-1)Cr for this case in an IIT maths text book
Are both the formule the one and the same....do they give the same answer whn tried for different combinations..

also tell if there is a direct formulae or something for the case when different balls
have to be put in similar boxes other than the method you have done..

also tell me when 7 different balls are put in to 3 different boxes
is the number of ways 3^7 or
is it 9!/(2!)

thanks in advance[/highlight:067b98f9b2]

First of all .... 9C7 is the same as 9C2. It doesn't matter which formula you use.nCr is the same as nC(n-r).

if 7 different ball are to be put n 3 different boxes, the answer is 3 ^ 7.That's because the first ball can go into any of the 3 boxes, the second into any of the 3 boxes.. and so on right till the seventh one.

Different formulae? I have no idea mate ... My humble two pennies worth of gyan, it is always better to steer clear of formulae when it comes to perm & comb problems. If you get the funda ... use logic and do it ... else leave it (in the exam i mean .. not now .. when you are preapring ) .. formulae are the perfect way of confusing perm comb problem... since you can literally derive a new on efor every problem.

And for the (2 2 0 0 0 ) case, .. no .. it is not 4!/(2!2!) .. though it first appears to be so ... do ine thing .. take four balls.. call them A, B, C, D and try making couples out of them You will find the second half is identical with the first half. That is because the BOXES are identical .... 4!/(2!2!) counts the same pair of balls in two boxes as different ... That is not so .. since the boxes are indistinguishable.


Hope that solves your doubts.

[catnapping]

well the total no. of ways in which the six sides of a cube(assuming the six sides are distinct and there is no circular prob associated and the six colours can be repeated) can be coloured using six coulours would be 6 raised to the power 6.

that doesn't seem right, does it?