Found 4 solutions by Edwin McCravy, greenestamps, ikleyn, mccravyedwin:
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!
There is a group of five children, where two of the children are twins. In how
many ways can I distribute 18 identical pieces of candy to the children, if the
twins must get 7 pieces of candy?
You didn't say "the twins must get 7 pieces of candy EACH" so I'll assume you
mean they get 7 pieces of candy together.
We put 7 pieces aside to give the twins. So we have 11 pieces to distribute to
the 3 non-twin children.
Suppose non-twin child A gets "a" pieces of candy, non-twin child B gets "b"
pieces of candy, and non-twin child C gets "c" pieces of candy. I will assume
everybody gets at least one piece of candy.
Then a+b+c=11, where a,b,c are all positive integers.
Let x=a-1, y=b-1, z=c-1, so a=x+1, b=y+1, c=z+1
a+b+c=11 so
(x+1)+(y+1)+(z+1)=11
x+y+z=8
So x, y, and z are, respectively, 1 fewer than the number of pieces of candy
that A,B, and C respectively get.
We make a row of 8 stars and 2 bars, for example ***|****|*. The number of
stars left of the leftmost bar, is x. The number of stars between the two bars
is y, and the number of bars right of the rightmost bar is z
For the example ***|****|*, the 3 stars left of the first bar means that x=3 and
a=4. The 4 stars between the bars mean z=4, and b=5. The 1 star on the end
means z=1 and c=2.
Take another example. ****||**** x=4, y=0, z=4, so a=5, b=1, c=5.
Take another example. ||******** x=0, y=0, z=8, so a=1, b=1, c=9.
There are 8 stars and 2 bars. The two bars are undistinguishable and the 8
stars are also undistinguishable. So there are 10!/(2!8!) = 45 ways to
distribute the 11 pieces of candy to the three non-twins.
Now we distribute the 7 pieces of candy to the twins. We could get the number of
ways to distribute with stars and only 1 bar. But it's easier just to realize
that the first twin can get 1 through 6 pieces and the second twin gets the
rest. So there are 6 ways to distribute the 7 pieces of candy to the twins.
So the answer is (45)(6) = 270 ways to distribute the 18 pieces of candy to
the 5 children.
Answer: 270
Edwin
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
The statement of the problem is not clear, leading to different possible interpretations. My interpretation differs from that shown in the response from tutor @Edwin in two ways. (1) The problem does not say that each child must get at least one candy, so the numbers each child gets are non-negative integers, not positive integers. (2) The number of candies each of the twins gets is not important; the only requirement is that together they get 7 pieces.
With that interpretation....
Take the 7 pieces of candy and give them to the twins (to be shared between them in any way).
That leaves 11 pieces of candy to be divided among the other three children. Using the "stars and bars" process, there are 11 stars (the pieces of candy) and 2 bars (to divided the stars (candies) into 3 groups). The number of ways to distribute the candies is then the number of ways of arranging the symbols
***********||
By a well-known counting principle, that number ways is
ANSWER (with this interpretation): 78
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
.
There is a group of five children, where two of the children are twins. In how
many ways can I distribute
18 identical pieces of candy to the children, if the twins must get 7 pieces of candy?
~~~~~~~~~~~~~~~~~~~~~
I came with third interpretation, which differs from that by Edwin's and @greenestamps.
I will assume that
(1) for twins and for other 5-2 = 3 children the distributions allow 0 pieces for some child/children
(but the sum of pieces for twins is always 7 and the sum of pieces for other three children is always 18-7 = 11)
(2) The number of pieces each twin gets is IMPORTANT.
So, my interpretation is half-way between that of Edwin and @greenestaps.
In support of this interpretation, I should say following:
(a) My assumption (1) is inhumane; but this problem is not about humanitarian mission - it is about
solving Math problem as it is given. So, I leave humanitarian reasons aside as irrelevant.
(b) My assumption (2) reflects the fact that the twins are two different distinguishable individuals;
therefore, when they obtain different numbers of pieces, it corresponds to different distinguishable distributions.
After all these explanations, below is the solution itself.
So, for 3 children, I have the "stars and bars" distribution problem with n=3 persons and k=11 pieces, with 0 (zero)
lower boundary of pieces for individuals.
For this 3 children, the number of possible distributions is = = = = 78.
For the twins, there are 7+1 = 8 different distributions (they correspond to numbers {0,1,2,3,4,5,6,7} of pieces
which one of the twins does obtain; then the other twin gets the rest of 7).
So, the ANSWER to the problem is this product 8*78 = 624.
This is my solution in my interpretation.
Answer by mccravyedwin(407) (Show Source): You can put this solution on YOUR website!
My interpretation is that if you 'distribute' candy among children, you certainly
don't allow the possibility of giving 11 pieces to one child and none to the
others. I wouldn't call that 'distributing' the candy. I guess I thought about
when I was teaching, when I would 'distribute' handouts or tests, I didn't leave
any students out.
I also disagree with Greenestamps and agree with Ikleyn that you just don't
consider twins as if they were one "double person", i.e., one person with two
bodies, i.e., indistinguishable. [BTW, is it "indistinguishable" or
"undistinguishable"?The spelling with the "i" sounds better.]
Anyway, all three of us used the same method, with our different
interpretations. Also, I prefer thinking through the stars and bars method
without plugging in partition formulas.
J
Edwin
RELATED QUESTIONS