Melody

Yes that is fine - that is what private messaging is for :)

Thank you Kim,  you have made my day  

Why don't you join up - We would love to have you here and it is VERY easy to join.  

You do not have to give an email address or anything.

Yes I am from Mars,  but I spend a lot of my Earthly time in Sydney, Australia.   LOL

Alan spends his Earthly time in England :)

hi Hawkeye

I am not here for long so you better make it quick !      

Thanks Nauseated,

I am actually wondering if my first answer - which I nutted out without a formula was also correct.

It started with a different premise - that a child could get no lollies, that may be the only reason it was 'wrong'.

Another for Mellie

In how many ways can we distribute 13 pieces of identical candy to 5 kids, if the two youngest kids are twins and insist on receiving an equal number of pieces?

Ok lets start with a different assumption this time.  Lets assume that every child gets at least one peice.

That means that 3 peices are gone to the elder 3 immediately.  Which means the most the little ones can get is 5 peices each

twins get 5 each      1 way

twins get 4 each, 5 peices left for 3 children     $${\left({\frac{({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{21}}$$    ways

twins get 3 each   7 left for 3 children       $${\left({\frac{({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)} = {\mathtt{36}}$$     ways

twins get   2 each   9 left for the 3 children     $${\left({\frac{({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}{{\mathtt{9}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{9}}){!}}}\right)} = {\mathtt{55}}$$     ways

twins get 1 each 11 left for the 3 older children       $${\left({\frac{({\mathtt{11}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}{{\mathtt{11}}{!}{\mathtt{\,\times\,}}({\mathtt{11}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{11}}){!}}}\right)} = {\mathtt{78}}$$

$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\small\textbf+\,}}{\mathtt{36}}{\mathtt{\,\small\textbf+\,}}{\mathtt{55}}{\mathtt{\,\small\textbf+\,}}{\mathtt{78}} = {\mathtt{191}}$$   ways  

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I still think my original answer might have been correct only it assumed some children might miss out altogether.   

I think

the number of ways that n identical votes can be distibuted to k people (assuming that each person gets at least one vote =  $$\binom{n+k-1}{n}$$

1.  4 students are running for club president in a club with 50 members. How many different vote counts are possible, if all 50 members are required to vote?

I am going to assume that everyone gets at least 1 vote.

(50+4-1)C50 = 53C50

$${\left({\frac{{\mathtt{53}}{!}}{{\mathtt{50}}{!}{\mathtt{\,\times\,}}({\mathtt{53}}{\mathtt{\,-\,}}{\mathtt{50}}){!}}}\right)} = {\mathtt{23\,426}}$$

2.  4 students are running for club president in a club with 50 members. How many different vote counts are possible, if members may choose not to vote?

I think this assumes that they all get at least 1 vote and at least 1 does not vote at all

(50+5-1)C50 = 54C50

$${\left({\frac{{\mathtt{54}}{!}}{{\mathtt{50}}{!}{\mathtt{\,\times\,}}({\mathtt{54}}{\mathtt{\,-\,}}{\mathtt{50}}){!}}}\right)} = {\mathtt{316\,251}}$$

If some people get no votes it would be higher - does it need to be worked out that way?

Ref:

Mellie,  Do you know what the correct answer is?

No anon, i guess you made a mistake   

Hi MG,

Why did you put those dollars signs in your code for?