I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if: (a) The candies are all different and the lunch boxes are all different? (b) The candies are all the same and the lunch boxes are all the same? (c) The candies are all the same and the lunch boxes are all different? (d) The candies are all different and the lunch boxes are all the same? (e) Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different?

Guest Mar 22, 2015

#9**+6 **

I see (a) a little differently......

I break this down into 2 tasks.......choose a piece of candy.....choose a luchbox to put it in....

For the first task ...we want to choose one of the 3 candies and one of the 4 lunchboxes to put it in = C(3,1) * C(4,1) = 12 ways to do this

So....after we've chosen the first piece of candy and the first lunchbox....we want to choose one of the 2 remaining pieces of candy and put it into one of the 3 remaining lunchboxes = C(2,1)*C(3,1) = 6 ways to do this

And finally, we just want to choose one of the two remaining lunchboxes to put the last piece of candy into....and there are 2 ways to do this.

So 12 ways x 6 ways x 2 ways = 144 ways

CPhill
Mar 24, 2015

#3**+5 **

Okay, you did not answer my question but I suppose it is same to assume you want to put the candy in the lunch boxes.

I have 3 pieces of candy to place in 4 lunch boxes.

In how many ways can I do this if:

(a) The candies are all different and the lunch boxes are all different?

Well you can put the first peice in any lunch box that is 4 choices, the same with the second 3rd so

it wll be 4*4*4 = 64 ways

(b) The candies are all the same and the lunch boxes are all the same?

1,1,1,0

2,1,0,0

3,0,0,0

That is 3 ways because the only thing that matters is how many lollies is in each box.

(c) The candies are all the same and the lunch boxes are all different?

3 from above * 4! = 3*4*3*2*1 I think (not really sure )

(d) The candies are all different and the lunch boxes are all the same?

3*3! maybe

(e) Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different?

3*4! * 3 I'm guessing. :))

Melody
Mar 24, 2015

#4**+5 **

The **correct** and **highly accurate** solutions presented below for your entertainment and edification . . . .

$$\;.\hspace{20pt} \text {a)\; } 4^3 = 64 \; \hspace{15pt}\leftarrow \hspace{15pt} \small \textcolor[rgb]{0,0,0}{\text {No restrictions:}}\\$$

$$\displaystyle \text {b)\; 3} \hspace{15pt}\leftarrow \hspace{15pt} \small \text {Enter "IntegerPartitions [3]" \; into \ Wolfram \ alpha.} \

\hspace{15pt} \text { This will show the values and the symbolic partitions.}\\

\;. \hspace{35pt}\tiny \text { (You may need to use your brain a little more than normal). }\\$$

$$\;. \hspace{15pt} \displaystyle \text {c)\; 20 } \hspace{15pt}\leftarrow \hspace{15pt} \left( {\begin{array}{*{20}c} Removed \ errant \ Binomial \\ \ from \ this \ post \\ \end{array}} \right) = \underbrace{\dfrac{(N+k-1)!}{k!}}_{Dist. \ box \ Indist. \ candy }\; \hspace{15pt}| \hspace{15pt} \Text {N=4 \; k=3} \\\\$$

$$\;. \hspace{15pt} \displaystyle \text {d)\; 7 } \hspace{15pt}\leftarrow \hspace{15pt} \binom {3}{3}\; +\; \binom {3}{2}\; +\; \binom {3}{1}\; =\; 7 \;

\text {(Indist. box; Dist. candy)}\\$$

$$\;. \hspace{15pt}\displaystyle \text {e)\; 40 } \hspace{15pt}\leftarrow \hspace{15pt} \left( \dfrac{4!}{2!}\right)* \left(2 \right) \;+\; \left( \dfrac{4!}{2!}\right) \;+\; \left(4\right)\\$$

$$\text {Reasoning:}\\

\displaystyle \text {All candy in 1 of 4 boxes (4)}\\

\displaystyle \text {2 combinations (XX Y\; \&\; XY X) \;distributed to boxes in (2)(4*3) = (2)(4!/2!) \;=\;(12*2) ways;}\\

\text {and 1 candy to each box distributed to boxes in (1)(4*3*2)/2 = 4!/2! = (12) ways.} \\

\text {Total 4+24+12 = 40 ways.}$$

All this plus or minus CDD . . .

$$\tiny \textcolor[rgb]{1,,0}{\text {(Corrected error in binomial formula for soultion c)}}\\$$

Nauseated
Mar 24, 2015

#5**0 **

But **Nauseated**, you are confusing me.

$$\;. \hspace{15pt} \displaystyle \text {c)\; 20 } \hspace{15pt}\leftarrow \hspace{15pt} \left( {\begin{array}{*{20}c} N \\ k \\ \end{array}} \right) = \underbrace{\dfrac{(N+k-1)!}{k!}}_{Dist. \ box \ Indist. \ candy }\; \hspace{15pt}| \hspace{15pt} \Text {N=4 \; k=3} \\\\$$

$$nCk = 4C3 = \frac{4!}{3!1!}=4$$

Melody
Mar 24, 2015

#6**0 **

A little CDD here. The **formula is correct**: (4+3-1)!/3! = (6!/3!); the **binomial is in error**.

Now **corrected and noted**.

Nauseated
Mar 24, 2015

#7**0 **

I don't know **Naus**, I still have a problem with this:

$$\;. \hspace{15pt} \displaystyle \text {c)\; 20 } \hspace{15pt}\leftarrow \hspace{15pt} \left( {\begin{array}{*{20}c} N+k-1 \\ k \\ \end{array}} \right) = \underbrace{\dfrac{(N+k-1)!}{k!}}_{Dist. \ box \ Indist. \ candy }\; \hspace{15pt}| \hspace{15pt} \Text {N=4 \; k=3} \\\\$$

$$^{(N+k-1)}C_k = \frac{(N+k-1)!}{k!(N+k-1-k)!}=\frac{(N+k-1)!}{k!(N-1)!}$$

That is still not your formula :/

Melody
Mar 24, 2015

#8**0 **

Technically, the equal sign made it true, but itâ€™s confusing: Remove errant Binomial from my mind and this post.

Nauseated
Mar 24, 2015

#9**+6 **

Best Answer

I see (a) a little differently......

I break this down into 2 tasks.......choose a piece of candy.....choose a luchbox to put it in....

For the first task ...we want to choose one of the 3 candies and one of the 4 lunchboxes to put it in = C(3,1) * C(4,1) = 12 ways to do this

So....after we've chosen the first piece of candy and the first lunchbox....we want to choose one of the 2 remaining pieces of candy and put it into one of the 3 remaining lunchboxes = C(2,1)*C(3,1) = 6 ways to do this

And finally, we just want to choose one of the two remaining lunchboxes to put the last piece of candy into....and there are 2 ways to do this.

So 12 ways x 6 ways x 2 ways = 144 ways

CPhill
Mar 24, 2015