Question

delete

Answer 1

FIRST THE COMMON CASE

ok so this would be a problem without replacement right? and we have three separate permutations (boxes where the order doesn't matter)

Answer 2

for the first box that would be 6P3, where we have
6:{b,b,b,w,w,e} (black,white,empty)

the order doesnt matter

Answer 3

the second box would only have 3 balls to choose from.

I'm not sure how to connect the second choose to the first one...

Answer 4

the second one should be 3P2, since we have 3 balls left and
we choose two from that. the three remaining balls depend on
the first choosing..

Answer 5

@Davidc
do you know how to calculate permutations like 6P3 ?

Answer 6

yes

Answer 7

very good

so we have first 6P3, then 3P2 and finally 1P1
for the permutation,order doesn't matter

Answer 8

permutation: order matter?
Combination: order doesn't matter?

Answer 9

ok I had them switched around

Answer 10

so we need combination within one box,right?
the order inside box doesn't matter

Answer 11

yes

Answer 12

6C3, 3C2 and 1C1

Answer 13

for example,
bbb we w is false
because "w" and "we" is alone

Answer 14

thats another question:

is wbb wb e allowed?

Answer 15

allow

Answer 16

OK because it is with a black ball and thus not alone

Answer 17

we should multiply 6C3, 3C2 and 1C1 with each other
because it's a sequence of events

Answer 18

but before we multiply,we must take away illegal configurations

Answer 19

the restriction of the white balls means there can't be a white ball in the third box.

if we have 1C1 with a white ball,thats not allowed............the 3C2 isn't a true 3C2, it makes it illegal to pick in such a way, that only a white ball is left......
this also depends on the first choosing.........................

Answer 20

usually the last ball can be:
-black
-white
-empty

now it can only be
-black
-empty

so one less. I wonder how to describe this in the right form

Answer 21

the last choice 1C1 isn't an actual choice, it depends from previous choices

Answer 22

there are 34 ways.
i just list all results.

Answer 23

:)

Answer 24

but i don't know how to cal

Answer 25

yeah ,its difficult.....

Answer 26

there are three illegal conditions:

1) white ball in last spot
2) one white ball in the middle spot

thats it.

Answer 27

the last choice irritates me the most. theres only one ball left.
the last choice purely depends on previous choices....

Answer 28

we have to subtract/remove the illegal situations

Answer 29

for 2) stop 1 alone ball in middle spot

I guess you can just take away two combinations from the problem?
e w
w e are the ones that are not allowed here.

Answer 30

once we have done that, we need to take away one other way -
b b w b _ w
b _ w b b w

Answer 31

b bb wwe
b ww bbe
b bw bwe
b be bww
e bb bww
e ww bbb
e bw bbw

Answer 32

we have to get rid of all combinations that end up with an illegal configuration:

so for example we need to take away

b b w b _ w
b _ w b b w ....mathematically.

this is like. ..maybe we should
calculate possible combinatins for the illegal result ?
and then subtract that

Answer 33

so I guess we could try to do this problem from the back to the beginning:

illegal combos
w (we) (anything 3) + w (any2) (any3) + (any1) (we) (any3)

Answer 34

first illegal choices:
w (we) (anything3) <- anything3 actually is depleted of w, we - can only be bbb

Answer 35

ok I think we don't really need the first one it is redundant.

w (we) (anything 3) + w (any2) (any3) + (any1) (we) (any3)

the second two already have the first one with it.

Answer 36

illegal combos
w (any2) (any3) + (any1) (we) (any3)

Answer 37

^those are the only combos we are concerned of

Answer 38

w (any2 )(any3) is depleted of w. so it will be:

1 x 5C2, 3C3 .

one time because only for w. only one chain of combinations
is illegal for this exception: when it starts with a white ball alone.

Answer 39

b (any2) (any3)
e (any2) (any3)

these are okay,so, we don't have them as an illegal combo

Answer 40

(any1) (we) (any3)

4C1 , 1 , 3C3

I have 4C1 because we can not choose we in this illegal scenario
they'r already "reserved" (otherwise wouldn't be the illegal sc.)

I have a "1" at we,because we or ew doesn't matter.

Answer 41

its a sequence, so they must be multiplied.

4C1 x 1 x 3C3

Answer 42

this is the amount of combinatins we lose to the we case

Answer 43

and

1 x 5C2 x 3C3

is the amount of combinations we lose to the [w] case

Answer 44

so, we would have

total combinations:
6C3x 3C2x 1C1

MINUS the illegal combinations:
4C1 x 1 x 3C3 + 1 x 5C2 x 3C3

Answer 45

but it is not true.

Answer 46

I already fear that :(

Answer 47

may be one illegal combination you did not consider

Answer 48

all combinations: 6C3x 3C2x 1C1

20 * 3 * 1 = 60

Answer 49

is that right ?

Answer 50

no

Answer 51

hmmm

Answer 52

it should be 34 ?

Answer 53

b bb wwe 3
b ww bbe 3
b bw bwe 12
b be bww 6
e bb bww 3
e ww bbb 1
e bw bbw 6
------------
34

Answer 54

how do you get the numbers for each row? :)

Answer 55

b1 b2b3 w1w2e
b2 b1b3 w1w2e
b3 b1b2 w1w2e

Answer 56

yeah,that's a really good point. in my calculation, i make a difference between
b1b2
b2b1
b3b1
b1b3

that shouldn't be the case though

Answer 57

i dont think it should be. black is black

Answer 58

b b b, w w, e
b b b, w e, w
b b w, b e, w
b b w, b w, e
b w w, b b, e
b w w, b e, b
w w e, b b, b

Answer 59

the balls are different

Answer 60

I understand

Answer 61

those are the only illegal combinations I see

w (any2) (any3)

(any1) (we) (any3)

Answer 62

(any1) (we) (any3)

there are actually two cases for this:

(any1) (w1e) (any3) and
(any1) (w2e) (any3)

Answer 63

also

w1 (any2) (any3)

w2 (any2) (any3)

Answer 64

4C1 x 2 x 1 = 8

2 x 10 x 1 = 20

Answer 65

60-28 = 32

Answer 66

two may be repeated

Answer 67

where repeated

Answer 68

i can't see

Answer 69

hmm.

Answer 70

I get the same numbers in the list