Question

Boxes numbered 1, 2, 3, 4, and 5 are kept in a row, and they are to be filled with either
a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. How
many different arrangements are possible, given that all balls of a given colour are
exactly identical in all respects?
(1) 8 (2) 10 (3) 15 (4) 22

Answer 1

any help!!

Answer 2

is 8 right? RBRBR,BRBRB,RRRRR (another 5 when B is arranged like BRRRR,RBRRR..etc)

Answer 3

what about BRBRR...

Answer 4

Yes,, comes to 10 now (another BRRRB)

Answer 5

brbrr,brrbr,brrrb,rbrbr,rbrrb,rrbrb

Answer 6

rrrrr
rrrrb,rrrbr,rrbrr,rbrrr,brrrr
brbrr,brrbr,brrrb,rbrbr,rbrrb,rrbrb
brbrb

Answer 7

the correct ans given is 22 !!

Answer 8

I don't see how that is possible

Answer 9

does it give a 'solution'

Answer 10

no, only the answ is given !!

Answer 11

I've listed all the possibilities....have no clue how they came up with 22.

Answer 12

:), thanks for your help

Answer 13

np...I'm still trying to figure out what they did wrong.

Answer 14

I found a 'solution' online

http://www.coolavenues.com/mbasp/snap.pdf

Each box can have either a red ball or blue ball, so total number of ways of filling are
\(2^5 = 32\). In adjacent positions, two blue balls can be filled in 4 ways, three blue in 3
ways, four blue in 2 ways, and five blue in 1 way. Total ways with blue balls in
adjacent positions = 10. So total number of ways where blue balls are not adjacent =
32 - 10 = 22.

but is is flawed...

there are 32 ways to arrange red and blue if you don't care about adjacency.
then you need to look at all the ones were they are adjacent

I agree with the 'two blue balls can be filled in 4 ways' and ' five blue in 1 way'

but the others are incorrect. which can be observed by looking at 'four blue in 2 ways'
bbbbr,bbbrb,bbrbb,brbbb,rbbbb which is 5 not 2

and there are 9 ways to arrange 3 blues where they are adjacent (not 3 as they state)

Answer 15

thanks, also it does consider the order factor !!!