Homework #4 — Math 173 — Spring ’09
Solutions
(1) 5.4; 6) Let the number of balls in the three boxes be x,yand z. We are told that x+y+z=
21. If zcounts the number of balls in the box with twice as many as the others combined,
then z= 2(x+y). Eliminating the z, this tells us 3(x+y) = 21 or x+y= 7. So one box
has 14 balls and the other two have {0,7},{1,6},{2,5}or {3,4}.
No matter which combination (e.g., {14,4,3}) we choose for the the number of balls
in each box, all three values are different. So we have 3! possible orderings in each case.
Since we are distributing distinct objects, for each ordering we can just use the multinomial
coefficient. Since the different orderings are different cases, these answers are then just
added up:
3! 21
14,7,0+21
14,6,1+21
14,5,2+21
14,4,3= 44651520.
An alternative solution is to realize that after the we choose the 14 balls for the one box
in 21
14ways, each remaining ball can go into one of two boxes with no restrictions. By the
MR, therefore, these last 7 balls can be placed in 27ways. Finally, we need to multiply by
3 to choose whether the box with 14 balls is first, second or third. So, this way, the answer
is 321
1427.
(2) 5.4; 14) First we convert this to a problem for which we’re looking for nonnegative solutions
by subtracting 4 from each side: (x1−1)+(x2−1)+(x3−1)+(x4−1) <96. As we saw in class,
we can deal with the less-than sign by adding an extra variable, x5. This variable absorbs
the difference between our minimum possible sum (zero) and our maximum possible (ninety
five). So we have as our modified equation (x1−1) + (x2−1) + (x3−1) + (x4−1) + x5= 95.
This is now a balls-and-boxes problem with r= 95 and n= 5. So the number of solutions
is 95+5−1
95 =99
4= 3764376.
It would also be natural to try to deal with the inequality before dealing with the “posi-
tive” requirement. Then, after adding the dummy variable you’d have x1+x2+x3+x4+x5=
99. But then you’d only want to subtract one from the first four variables as the problem
requires that they be positive, while x5is certainly allowed to be zero. This leads to the
same final equation we obtained above.
(3) 5.4; 16) In the first step, person one is given one of 10 books, say Don Quixote. Then
persons two and three get books. Then the remaining 7 books are distributed. Suppose the
last of these 7 is an advance copy of Pride & Prejudice & Zombies that goes to the first
person. Alternatively, we could have made the same choices except have Pride & Prejudice
& Zombies be given to the first person in step one and Don Quixote be given as the last
book. These two cases are counted differently, however they lead to the same ultimate
distribution of books and therefore should only be counted once. In the language of the
book, this strategy violates the Set Composition Principle.
Another error is in how many ways there are to distribute those final 7 books. rdistinct
objects distributed to ndistinct boxes can be distributed in nrpossible ways. So the number
of possibilities here is 37, not 73as the book lists.
(4) 5.4; 20) The number of possible games for the first bag is 30
20. The number of possible toys
for the second bag is 20
15. The number of possible presents for the third bag is 50−20−15
15 .
These choices are all independent, so by the MR, the answer is 30
2020
1515
15= 465817912560.
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