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Math 55 HW 8 solutions
Chapter 6.5
6.5: 10, 11, 16, 18, 20, 24, 26x, 47.
10 (a) Since there are 6 kinds of croissant, this is the same as counting the number of non-negative integer
solutions to x1 + x2 + x3 + x4 + x5 + x6 = 12. That is, there are C(12 + 6 − 1, 6 − 1) = C(17, 5) possbile
choices.
(b) Same as before, but for solutions to x1 + x2 + x3 + x4 + x5 + x6 = 36, of which there are C(41, 5).
(c) Now we count the number of solutions to x1 + x2 + x3 + x4 + x5 + x6 = 24 where xi ≥ 2. Letting
yi = xi − 2, this is the same as counting non-negative integer solutions to y1 + y2 + y3 + y4 + y5 + y6 = 12.
So the answer is C(17, 5).
(d) Now we count solutions to x1 + x2 + x3 + x4 + x5 + x6 = 24 where x6 ≤ 2. We do this by counting all
non-negative solutions and then throwing away (subtracting) the solutions for which x6 > 2. In detail:
There are as many solutions to x1 + x2 + x3 + x4 + x5 + x6 = 24 with x6 > 2 as solutions to x1 + x2 + x3 +
x4 + x5 + x6 = 22 with xi ≥ 0, that is C(27, 5). We subtract this from C(29, 5) to get C(29, 5) − C(27, 5)
ways to choose two dozen croissants with no more than two broccoli.
(Note: It is also possible to count and add the possibilities with no, one and two broccoli croissants.
This would be just slightly more work and should give the same answer.)
(e) Now we count solutions to x1 + x2 + x3 + x4 + x5 + x6 = 24 with x3 ≥ 5 and x4 ≥ 3. Let y3 = x3 − 5 and
y4 = x4 − 3. Then this count is the same as counting the solutions to x1 + x2 + y3 + y4 + x5 + x6 = 16
with non-negative numbers. So, C(21, 5).
(f) This is equivalent to counting solutions to
x1 + x2 + x3 + x4 + x5 + x6 = 24 − 1 − 2 − 3 − 1 − 2 = 15
where x6 ≤ 3.
There are C(20, 5) to this equation without the constraint, and there are as many solutions with x6 ≥ 4
as there are unconstrained solutions to
x1 + x2 + x3 + x4 + x5 + x6 = 11,
so C(16, 5).
Taking the difference, there are C(20, 5) − C(16, 5) ways to choose the croissants.
11 Solution in back of book.
16 (a) Since each xi is at least 2, if we yi = xi − 2 we get non-negative yi and the equation becomes
y1 + y2 + y3 + y4 + y5 + y6 = 17.
This has C(22, 5) solutions.
(b) Now let y1 = x1 − 1, y2 = x2 − 2, y3 = x3 − 3, y4 = x4 − 4, y5 = x5 − 5, y6 = x6 − 6. We want to count
solutions to
y1 + y2 + y3 + y4 + y5 + y6 = 29 − 1 − 2 − 3 − 4 − 5 − 6 = 8,
of which there are C(13, 5).
(c) First we count the number of solutions with x1 ≥ 6 (note: because we are assuming x1 to be an integer,
x1 ≥ 6 is an appropriate form of the logical negation to x1 ≤ 5.). This is the same as the number of
non-negative solutions to
y1 + x2 + x3 + x4 + x5 + x6 = 23,
that is, C(28, 5).
Then we subtract this from the number of solutions to the original equation. The answer is C(34, 5) −
C(28, 5).
(d) We now must combine the tricks from the previous items: By letting y2 = x2 − 9, we see that we want
to count the number of solutions to
x1 + y2 + x3 + x4 + x5 + x6 = 20
for which x1 < 8.
Now we count the complimentary set, so solutions to
x1 + y2 + x3 + x4 + x5 + x6 = 20
for which x1 ≥ 8. Letting y1 = x1 − 8, we see that this is the same as the number of solutions to
y1 + y2 + x3 + x4 + x5 + x6 = 12,
which is C(17, 5).
Now we have to not forget that we just counted the compliment of the set we actually care about. The
answer is C(25, 5) − C(17, 5).
18 This is the same as counting the number of ways to put 20 distinct objects (each decimal place) into 8
distinguishable boxes (the choice of digit for said decimal place) with ni objects in the ith box, where
n1 = 2, n2 = 4, n3 = 3, n4 = 1, n5 = 2, n6 = 3, n7 = 2, n8 = 3.
Theorem 4 tells us that there are
n!
2!4!3!2!3!2!3!
such choices.
20 Given a solution to
x1 + x2 + x3 ≤ 11
we let x4 = 11 − x1 − x2 − x3 . Then we get a solution to the equation
x1 + x2 + x3 + x4 = 11.
Conversely, given such a solution, it will certainly hold that
x1 + x2 + x3 ≤ 11.
Thus the 2 sets (of solutions to each equation) are in bijection. The latter set has C(11 + 4 − 1, 3) = C(14, 3)
elements.
24 This is an application of theorem 4.
15!
.
1!2!3!4!5!
26x We count the number of numbers of the form x1 x2 x3 x4 x5 x6 so that exactly one of the xi = 9 and x1 + x2 +
x3 + x4 + x5 + x6 = 13. There are 6 cases: x1 = 9, x2 = 9, etc.
Conside the case that x6 = 9. Then we want to count solutions to x1 + x2 + x3 + x4 + x5 + 9 = 13, which is
the same as solutions to x1 + x2 + x3 + x4 + x5 = 4. There are C(8, 4) such solutions. (Note that because
the remaining xi are supposed to be digits that aren’t 9, so we should properly be counting solutions so that
xi ≤ 8. However, because they add up to 4, this constraint never matters.)
Since the other 5 cases are the same, there are 6 × C(8, 4) such numbers.
47 Solution in back of book.
Chapter 7.1
7.1: 6, 18, 19, 20, 34, 36, 38.
6 In a deck of cards the set of aces has 4 elements and the set of hearts has 13 elements. The only card in
both sets is the ace of hearts. By the inclusion-exclusion principal there are 4 + 13 − 1 = 16 cards which are
an ace or a heart. Thus the probability of picking such a card is 16/52 = 4/13.
18 In each suit it is easy to count that there are 10 possible straight flushes (one starting at each of A,
2, 3, . . . , 10.). Thus there are a total of 40 = 10 × 4 straight flushes. There are of course C(52, 5) possible 5 card hands, so the probability of a straight flush is 40/C(52, 5).
19 Solution in back of book.
20 There are only 4 possible royal flushes, so the probability of getting one is 4/C(52, 5).
34 (a) If no one can win twice then there are 50 × 49 × 48 × 47 ways to draw 4 winners in order, and only 1
1
way for them to be Bo, Colleen, Jeff, and Rohini in that order. So the probability is 50×49×48×47
.
(b) If people can win more than once then there are 504 ways to draw 4 winners. So the probability
decreases to 5014 .
36 When 2 dice are rolled there are 5 ways to make 8: 2+6, 3+5, 4+4, 5+3, 6+2. (Note that rolling a 2 and
then a 6 is different from rolling a 6 and then a 2, so we have to make sure to count both.). Thus the
probability of rolling an 8 is 5/36.
When 3 dice are rolled there are 20 ways to make 8: 1+1+6, 1+2+4, 1+3+3, 1+4+2, 1+5+1, 2+1+5,
2+2+4, 2+3+3, 2+4+2, 2+5+1, 3+1+4, 3+2+3, 3+3+2, 3+4+1, 4+1+3, 4+2+2, 4+3+1, 5+1+2, 5+2+1,
6+1+1. Since there are 63 ways to roll 3 dice, this means a probability of 20/63 of rolling an 8 with 3 dice.
This is smaller than 5/36.
38 There are 23 = 8 possible ways to flip a coin 3 times.
(a) E1 = {thh, tht, tth, ttt} has 4 elements, so a probability of 1/2. E2 = {hhh, hht, thh, tht} has 4 elements,
so a probability of 1/2. E1 ∩E2 = {thh, tht} has 2 elements, so a probability of 1/4. Since 1/2×1/2 = 1/4
these events are independent.
(b) E1 = {thh, tht, tth, ttt} and E2 = {hht, thh}, so E1 ∩ E2 = {thh}. Thus
P (E1 ∩ E2 ) = 1/8 = 1/2 × 1/4 = P (E1 )P (E2 )
and the events are independent.
(c) E1 = {hth, htt, tth, ttt} and E2 = {hht, thh}, so E1 ∩ E2 = φ. Thus
P (E1 ∩ E2 ) = 0 6= 1/2 × 1/4 = P (E1 )P (E2 )
and the events are not independent.