Permutation and Combinations Help (1 Viewer)

[AlexMD]

How many ways are there to split 4 red, 5 blue and 7 black balls among:

i) 2 boxes, without any restrictions?

i) 2 boxes,with no box empty?

If you answer it could you please explain how you got to your answer as well? Been stuck on it for a while now...would really appreciate the help

 

[seanieg89]

Each arrangement corresponds to a triple of non-negative integers (r,u,b) the number of red, blue, and black balls respectively in box A. Counting these triples is an easy application of the rule of product, keeping in mind the possible values these integers can take.
The second part follows by considering the number of arrangements WITH an empty box and subtracting this from your first answer.

i) 5 x 6 x 8 = 240

ii) 240 - 2 = 238

 

[k02033]

Assuming the boxes are distinct,

(write out all the possibilities for arranging 4 red balls and then 5 blue and you will see a pattern)

There are 5 ways of putting 4 objects into 2 boxes, 6 ways of putting 5 objects into 2 boxes so 8 ways ways of putting 7 objects into 2 boxes. So 5*6*8=240, if boxes are not distinct divide by 2, ie 120

Only 2 ways of having one of the boxes empty, either box 1 empty or box 2 empty, so 240-2=238. if boxes not distinct 120-1=119.

 

[nikkifc]

Why is this in Ext 2?

 

[AlexMD]

Why is this in Ext 2?

It was in an Extension 2 Paper and I'm pretty sure it's considered harder 3 unit?


Hope I didnt cause any confusion or distress...