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# 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 equ

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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?

Apr 30, 2015

#10
+9

Hi Melody and Mellie,

For this question  the answer is

$$\Displaystyle \binom{15}{2}+\binom{13}{2}+\binom{11}{2}=238$$

Nauseated

May 3, 2015

#1
+4

I don't know Mellie but I have a theory  lol

If the youngest 2 children get 6 each then there is 1 peice left that can got to any one of the other 3 children.

so there is 3 ways this can happen  3 ways

If the youngest 2 children get 5 peices each then the other 3 peices can be allocated as

(1,1,1 ) [1way]     (1,2,0 ) [3!=6ways]   or    (3,0,0) [3ways]   = 10 ways

If the youngest 2 children get 4 peices each then this is how the other 5 peices can be allocated

(5,0,0) [3!/2!=3 ways] (4,1,0) [3!=6 ways]  (3,2,0) [6ways]  (3,1,1)  [3ways]  (2,2,1) [3ways]  =  21 ways

If the youngest 2 children get 3 peices each then this is how the other 7 peices can be allocated

(7,0,0) 3ways (6,1,0) 6 ways(5,2,0) 6 ways (5,1,1) 3 ways (4,3,0)6 ways (4,2,1) 6 ways (3,3,1) 3 ways (3,2,2) 3 ways   =   3+6+6+3+6+6+3+3 = 36 ways

Now I am going out on a ledge and say that I can see a pattern forming here

3 = 3C3        10= 5C3        21= 7C3       36= 9C3        .............

So maybe if the 2 youngest get 2 peices each then there will be     11C3 = 165 ways to distribute the other 9 peies

and maybe if the 2 youngest get 1 peice each there will be   13C3 = 286 ways to distribute the other 11 peices between the other 3 children.

and if the yougest 2 get none each then there will be 15C3 = 455 ways to distribute the other 13 peices between the other 3 children.

Total = $${\mathtt{455}}{\mathtt{\,\small\textbf+\,}}{\mathtt{286}}{\mathtt{\,\small\textbf+\,}}{\mathtt{165}}{\mathtt{\,\small\textbf+\,}}{\mathtt{36}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{976}}$$    ways altogether. --------------------------------

Mmm

If this is correct, I have found that n identical peices of candy can be distribued

between 3 children $$^{(n+2)}C_3$$     It would be good if someone could walk through the logic of why this is so :/

I shall think upon it May 1, 2015
#2
+6

This is incorrect, but they gave me a hint:

Hint(s): Deal with the restriction first.

May 1, 2015
#3
0

I did deal with the restrictions first :)

May 2, 2015
#4
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I'm not sure sorry. I don't know what else to do. :/

May 2, 2015
#5
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I didn't ask a question Mellie, you are confusing me :/

May 3, 2015
#6
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Yeah, I know you didn't ask a question, it's just that I don't know what other procedures we could follow to get the correct answer

May 3, 2015
#7
0

Mellie,  Do you know what the correct answer is?

May 3, 2015
#8
+4

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. May 3, 2015
#9
0

No, sadly, I don't know the solution and this is incorrect. I'm so sorry for you having to take so much time to solve my problems :(

May 3, 2015
#10
+9

Hi Melody and Mellie,

For this question  the answer is

$$\Displaystyle \binom{15}{2}+\binom{13}{2}+\binom{11}{2}=238$$

Nauseated

Guest May 3, 2015
#11
+5

Hi Mellie,

Your problems are time consuming and frustrating but I really like struggling with probablity questions.

I think the other mathematicians do too.

You do not need to appologise. Ok Nauseated - we will be waiting May 4, 2015
#12
+2

238 is not correct either. The answer is 308. You have to do combinations for each possibility and then add all of the combinations. The sum of the combinations is 308.

Aug 9, 2016
#15
+6

Naus’ solution dàmn well is correct!

I don’t know why Naus didn’t post a follow up with the explanation for the solution here, but he did send it to me in a tutoring session several weeks before this was posted. I’m sure it’s the exact same question that is here. I remember his answer was correct—I’d remember better if it wasn’t.  I’ll find his solution and comments and post them here.

GingerAle  Aug 11, 2016
#13
0

@Mellie

Are you taking AoPS online courses? You shouldn't cheat like this!

Aug 11, 2016
#16
+1

If you were really a good Detective then you’d know the “S” in AoPS stands for “Solving.”  Millie solved the problem of learning complex mathematics by coming here. In no way is she “cheating” by doing this.

Millie is very intelligent. Here is a very rare compliment from our resident troll, Nauseated.

http://web2.0calc.com/questions/hi-sorry-im-new-here-and-dont-know-what-to-do-now#r36

The compliment is sincere. Believe me, Naus doesn’t need to compliment someone to troll someone else. Most everyone on here will know that.

When this was posted, Millie was in 7th grade. I remember when I was in 7th grade, I didn’t even know what a combination was, but I could make change for a dollar, or a Euro, and sometimes a British Pound.   At least as long as my calculator had a good battery, I was on the money.  I was very proud of my skill. GingerAle  Aug 11, 2016
#14
+1

I'd say:

c(15,2)+c(13,2)+c(11,2)+c(9,2)+c(7,2)+c(5,2)+c(3,2)

=308

Aug 11, 2016
edited by Detective  Aug 11, 2016
#17
+6

I’d say:

I’d say:

Your detective skills suck, Detective.  I know because I went to detective school where I graduated with a “D minus.”  They taught me to keep two loaded magnums in my purse. One loaded with booze and the other with bullets.  That’s why I have five slugs in me: four are bourbon and one is a .44.

The minus on the “D” came from breaking a cardinal rule: never mix your booze with bullets.  I was lucky it was only a minus: others have had to do “Hail Marys” for breaking cardinal rules. they were really F'd.  Even with my “D minus” I’m an OK private eye. I get a couple of cases a month and they pay a couple of monthly bills –one from my probation officer and the other from my bookie.

When I socialize with my fellow chimps, it’s a banana daiquiri, when I’m on a case, it’s a bourbon on the rocks.  I’m on my sixth case now; I solved and finished the last five cases. Every bottle that isn’t broken is polished and displayed on my trophy shelf. These trophies are part of my pride, but not my joy –empty bottles always seem to lack joy.

This morning, while enjoying the quiet night watches with Bogart, my cat, someone kicked my door open and barged in. It was a wanabe detective.  He needed certification for his private díck license. He presented his credentials and the paperwork for his current case. He started to show my his private díck, but I told him to zip it.  I thought this might be a hard case, but as poor as I am at math, a quick perusal still told me all I needed to know.

I asked him, “Did you take a harsh laxative today?” He took too long to answer, so I continued, “I’d say: you did.  Unless you have a better reason for this cráp that makes you look like a third-rate mathematician and a fourth-rate detective.”

My cat and I are still waiting for his reply.   . . .

GingerAle  Aug 11, 2016
#18
0

This is a d**n funny parody of the old noir private detective films and novels. You must be a devoted fan of the genre to write something like this.  When I was growing up I had two dogs named Boggy and Cagney after Humphrey Bogart and James Cagney.

Are you Nauseated’s real life girlfriend?

Aug 12, 2016
#19
+2

This an AoPS Homework Question. This can be done entirely by casework.

If done correctly, you should have 105+78+55+36+21+10+3=308

Dec 2, 2016