A box contains 1 red cube, 2 identical blue cubes and 2 identical green cubes. What is the number of distinct ways one can take all the cubes out of the box so that all blue cubes are removed before all green cubes are removed?
In Greg’s explanation for this-
1st- he puts G at the end i.e (R B B G) G and does 4!/2! = 12
2nd - he puts R at the end i.e (B B G) G R and does 3!/2! = 3
Total = 15
My doubt - the question clearly says that the all blue cubes are removed before all green cubes. Doing 4!/2! or 3!/2! for (RBBG) and (BBG) respectively has possibilities where the G precedes the B
Example- (BGBR) G or (BGB) GR to list some.
This is a contradiction to what the question is asking. Let me know if I’m right or making a stupid/silly mistake in interpreting this whole thing.
You’re correct. Greg’s method does not guarantee that all blue cubes are removed before all green cubes.
The correct approach would be to consider the two blue cubes and two green cubes as a single group, with the condition that the blue cubes must come before the green cubes. This gives us
{4 \choose 2 }= 6
distinct arrangements for the blue and green cubes.
The red cube can be placed in any of the 5 positions in this sequence (before, between, or after the blue-green group), giving us a total of
5×6=30
distinct ways to remove all the cubes from the box.
or
To satisfy the condition that all blue cubes (B) are removed before all green cubes (G), we can have the following scenarios:
BB comes before GG
BB comes between the two Gs
BB comes before the first G
The total number of distinct sequences can be found by combining these scenarios.
BBGG + (R): We can insert R (red cube) in any of the positions in the sequence BBGG. We have 5 positions (BBGG, B_BGG, BB_GG, BBG_G, BBGG) to place R. So we have 5 sequences for this scenario.
BGBG + (R): Again, we have 5 positions (BGBG, B_GBG, BG_BG, BGB_G, BGBG) to insert R. So, we get 5 sequences for this scenario as well.
BBG + (G) + (R): Now, we insert the second G in the sequence. We have 4 positions (BBG, BB_G, BBG, BBG__) to insert G. For each of these sequences, we then have to insert R, which gives us 4 * 5 = 20 sequences.
Adding up all the possibilities: 5 (from scenario 1) + 5 (from scenario 2) + 20 (from scenario 3) = 30
