A couple counting problems....(full work shown..)

[Daniel_Feldman]

Just want to check and see if my logic/thought process is right for each of these...

1) You have 4 blocks, each with a letter on it (A, B, C, D). You also have 2 boxes, one red and one blue. You want to divide the 4 bocks evenly among the 2 boxes (i.e., 2 blocks per box).
a) How many ways could you do this?
b) What if both boxes were red?
c) What if you had 6 blocks (labeled A-F) and 3 boxes (red, blue, yellow)?
d) What if all 3 blocks were one color?

My thought process:

a) For the first box, we choose 2 blocks from 4. For the second box, it's 2 from 2. Then, we have 2! options for the choice of boxes (since there are 2 different boxes). So the answer I got is 4C2 x 2C2 x 2!=12 ways

b) If both boxes are the same, we simply divide the above answer by 2!, and get 6 ways.

c) Using the same idea as in part (a), we choose 2 from 6, then 2 from 4, then 2 from 2, and 3! possibilities for the boxes, so it's 6C2 x 4C2 x 2C2 x 3!=540 ways.

d) Dividing by 3! yields 90 ways.



2. We toss a die twice. A is the event that at least one of the numbers is 3. B is the event that the sum of the numbers is 7. C is the event that both numbers are odd.

a) Are A and B independent?
b) Are A and B mutually exclusive?
c) Are A and C independent?
d) Are A and C mutually exclusive

My logic:

a) We need to check whether P(AB)=P(A)P(B). P(A) is 11/36 [(1,3), (3,1), (2,3), (3,2), etc..but only one option for (3,3)]. P(B)=1/6 (6 ways to get sum of 7, 36 possible outcomes). P(AB), I think, is 1/18, because we want both A and B, and of all the combinations that yield a sum of 7, only 2 have at least one 3, so P(AB)=2/36=1/18. Since P(AB) doesn't equal P(A)P(B), they are not independent.

b) Mutually exclusive means P(A U B)=P(A)+P(B). P(A U B) is where I'm confused. How do you work with the fact that (3,4) and (4,3) lie in both A and B?

c) P(C)=1/4 and P(AC) =5/36. Doing the multiplication shows that they aren't independent.

D) Same question about P(A U C).


Thanks for the help.

 

[pka]

Let’s answer in general. Suppose that we have N different letters and K different colored boxes. (N must be a multiple of K, such as 6=2(3))
The answer to part a) is \(\displaystyle \frac{{N!}}{{\left( {K!} \right)^{\frac{N}{k}} }}\).

For part b) in which all the boxes are the same then the answer is
\(\displaystyle \frac{{N!}}{{\left[ {\left( {K!} \right)^{\frac{N}{k}} } \right]\left( {\frac{N}{k}} \right)!}}\)

 

[Daniel_Feldman]

I'm not so sure. By your formula, if we have 2 (different) boxes and 4 letters, the number of ways of dividing them is 4!/(2!)^(4/2)=24/(2)^2=6 ways. However, if we have A, B, C, and D and red and blue boxes, then we can have AB, AC, AD, BC, BD, or CD, which is 6 different options, and we have that for both the red and the blue boxes so that would yield 12 total ways.

 

[pka]

I'm not so sure. By your formula, if we have 2 (different) boxes and 4 letters, the number of ways of dividing them is 4!/(2!)^(4/2)=24/(2)^2=6 ways. However, if we have A, B, C, and D and red and blue boxes, then we can have AB, AC, AD, BC, BD, or CD, which is 6 different options, and we have that for both the red and the blue boxes so that would yield 12 total ways.

\(\displaystyle \begin{array}{*{20}c} {red} &\vline & {blue} \\\hline {A,B} &\vline & {C,D} \\ {A,C} &\vline & {B,D} \\ {A,D} &\vline & {B,C} \\ {B,C} &\vline & {A,D} \\ {B,D} &\vline & {A,C} \\ {C,D} &\vline & {A,B} \\ \end{array}\)

That is six ways.
Where do you get 12?

 

[Daniel_Feldman]

My bad. Makes sense now. Thanks.

So basically, in the case where both boxes are identical, we need to divide by 2! since AB in box 1 is the same as AB in box 2, right?