I really need help please! I have no idea how to do it. Sorry!!!

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

In how many ways can you do what?

it is the different parts of the questions. Please help!!

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.                      :))

 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 . . .

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

But Nauseated, you are confusing me.

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

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

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

Now corrected and noted.

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  :/

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