Transcript Math 106 – Exam #2 - Review Problems 1. Fourteen different candy bars (Snicker’s, Milky Way, etc.) are available for a mother.

Math 106 – Exam #2 - Review Problems
1.
Fourteen different candy bars (Snicker’s, Milky Way, etc.) are available for a mother to give
to her daughter.
(a) How many ways can the mother give the daughter exactly four of the candy bars?
C(14,4) = 1001
(b) How many ways can the mother give the daughter exactly four or five of the candy bars?
C(14,4) + C(14,5) = 1001 + 2002 = 3003
(c) How many ways can the mother give the daughter more than three but less than eight of
the candy bars?
C(14,4) + C(14,5) + C(14,6) + C(14,7) = 1001 + 2002 + 3003 + 3432 = 9438
(d) How many ways can the mother give the daughter more than ten of the candy bars?
C(14,11) + C(14,12) + C(14,13) + C(14,14) = 364 + 91 + 14 + 1 = 470
(e) How many ways can the mother give the daughter at least ten of the candy bars?
C(14,10) + C(14,11) + C(14,12) + C(14,13) + C(14,14) = 1001 + 364 + 91 + 14 + 1 = 1471
(f) How many ways can the mother give the daughter less than six of the candy bars,
including the possibility of giving none?
C(14,0) + C(14,1) + C(14,2) + C(14,3) + C(14,4) + C(14,5) = 1 + 14 + 91 + 364 + 1001 + 2002
(g) How many ways can the mother give the daughter at most six of the candy bars, = 3473
including the possibility of giving none?
C(14,0) + C(14,1) + C(14,2) + C(14,3) + C(14,4) + C(14,5) + C(14,6) = 6476
(h) How many ways can the mother give the daughter some but not all of the candy bars?
214 – 2 = 16,382
(i) How many ways can the mother give the daughter more than three of the candy bars?
214 – [C(14,0) + C(14,1) + C(14,2) + C(14,3)] = 214 – 470 = 15,914
(j) How many ways can the mother give the daughter at most 11 of the candy bars,
including the possibility of giving none?
214 – [C(14,12) + C(14,13) + C(14,14)] = 214 – 106 = 16,278
2.
Consider the equation x + y + z = 3 .
(a) How many solutions in non-negative integers are there?
5!
—— = 10
2! 3!
(b) Write each possible solution in non-negative integers together with either a sequence of
slashes and “c”s or a sequence of “u”s and “b”s to represent that solution.
x = 3, y = 0, z = 0 c c c / /
x = 2, y = 0, z = 1 c c / / c
x = 0, y = 2, z = 1 / c c / c
x = 0, y = 3, z = 0 / c c c /
x = 1, y = 2, z = 0 c / c c /
x = 0, y = 0, z = 3 / /c c c
x = 1, y = 0, z = 2 c / / c c
x = 2, y = 1, z = 0 c c / c /
x = 1, y = 1, z = 1 c / c / c
x = 0, y = 1, z = 2 / c / c c
3.
Fourteen identical candy bars (all Snicker’s) are available for second-grade teacher Rachael
to distribute either to the nine other teachers in the lounge or to the 21 second-graders in her
classroom.
(a) How many ways can Rachael distribute the candy bars among the other teachers in the
lounge excluding herself?
22!
x1 + x2 + … + x8 + x9 = 14
——— = 319,770
8! 14!
non-negative integers
(b) How many ways can Rachael distribute the candy bars among the other teachers in the
lounge and herself?
23!
x1 + x2 + … + x9 + x10 = 14
——— = 817,190
9! 14!
non-negative integers
(c) How many ways can Rachael distribute the candy bars among the second-graders in her
classroom?
34!
x1 + x2 + … + x20 + x21 = 14
——— = 1,391,975,640
20! 14!
non-negative integers
(d) How many ways can Rachael distribute the candy bars among the second-graders in her
classroom and herself?
x1 + x2 + … + x21 + x22 = 14
non-negative integers
35!
——— = 2,319,959,400
21! 14!
(e) How many ways can Rachael distribute the candy bars among the other teachers in the
lounge, if she keeps exactly one for herself and guarantees that each teacher in the lounge
gets at least one?
12!
x1 + x2 + … + x8 + x9 = 13
x1 + x2 + … + x8 + x9 = 4
——— = 495
8! 4!
positive integers
non-negative integers
(f) How many ways can Rachael distribute the candy bars among the second-graders in her
classroom, if she keeps exactly one for herself and guarantees that each second-grader
gets at least one?
zero (0), since this is impossible to do
(g) How many ways can Rachael distribute the candy bars among the other teachers in the
lounge excluding herself, if she identifies one teacher in the lounge who will get at least
five?
17!
x1 + x2 + … + x8 + x9 = 14
x1 + x2 + … + x8 + x9 = 9
——— = 24,310
8! 9!
5  x9
non-negative integers
(h) How many ways can Rachael distribute the candy bars among the second-graders in her
classroom, if she identifies one second-grader in her classroom who will get at most six?
x1 + x2 + … + x20 + x21 = 14
x1 + x2 + … + x20 + x21 = 14
x21  6
7  x21
x1 + x2 + … + x20 + x21 = 7
non-negative integers
27!
——— = 888,030
20! 7!
1,391,975,640 – 888,030 = 1,391,087,610
3.-continued
(i) How many ways can Rachael distribute the candy bars among the other teachers in the
lounge excluding herself, if she identifies one teacher in the lounge who will get at least
three and identifies another teacher in the lounge who will get at least two?
(ANSWERS ON NEXT SLIDE)
(j) How many ways can Rachael distribute the candy bars among the second-graders in her
classroom, if she identifies one second-grader in her classroom who will get at least four
and identifies another second-grader in her classroom who will get at most two?
(ANSWERS ON NEXT SLIDE)
(k) How many ways can Rachael distribute the candy bars among the other teachers in the
lounge excluding herself, if she identifies one teacher in the lounge who will get at least
four but at most eight?
(ANSWERS ON NEXT SLIDE)
(l) How many ways can Rachael distribute the candy bars among the second-graders in her
classroom, if she identifies one second-grader in her classroom who will get at least four
and identifies another second-grader in her classroom who will get at most eight?
(ANSWERS ON NEXT SLIDE)
3.-continued
(i)
x1 + x2 + … + x8 + x9 = 14
3  x8 & 2  x9
(j) x1 + x2 + … + x20 + x21 = 14
4  x20 & x21  2
x1 + x2 + … + x8 + x9 = 9
non-negative integers
x1 + x2 + … + x20 + x21 = 10
x21  2
17!
——– = 24,310
8! 9!
x1 + x2 + … + x20 + x21 = 10
3  x21
30!
27!
——— = 30,045,015
——— = 888,030
20! 10!
20! 7!
30,045,015 – 888,030 = 29,156,985
(k) x1 + x2 + … + x8 + x9 = 14
4  x9  8
x1 + x2 + … + x8 + x9 = 10
x9  4
x1 + x2 + … + x8 + x9 = 10
5  x9
18!
13!
——— = 43,758
——— = 1287
8! 10!
8! 5!
43,758 – 1287 = 42,471
(l) x1 + x2 + … + x20 + x21 = 14
4  x20 & x21  8
x1 + x2 + … + x20 + x21 = 10
x21  8
x1 + x2 + … + x20 + x21 = 10
9  x21
30!
21!
——— = 30,045,015
——— = 21
20! 10!
20! 1!
30,045,015 – 21 = 30,044,094
(m) How many ways can Rachael distribute the candy bars among the second-graders in her
classroom, if she identifies one second-grader in her classroom who will get at least four
x1 + x2 + … + x20 + x21 = 14
x1 + x2 + … + x20 + x21 = 14
but at most ten?
4  x21
4  x21 & 11  x21
x + x + … + x + x = 14
1
2
4  x21  10
20
21
30!
23!
——— = 30,045,015
——— = 1771
20! 10!
20! 3!
30,045,015 – 1771 = 30,043,244
4. A child likes two kinds of candy bars: Milky Way bars and Snicker’s bars.
(a) How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar,
be arranged in a row?
220 = 1,048,576
(b) How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar,
be arranged in a row so that candy bars of the same type are never adjacent?
2
(c) How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar,
be arranged in a row so that Milky Way bars are never adjacent?
F(20) = 17,711
4.-continued
(d) How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar,
be arranged in a row so that Snicker’s bars are never adjacent?
F(20) = 17,711
(e) How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar,
be arranged in a row so that at least two Milky Way bars are adjacent?
220 – F(20) = 1,048,576 – 17,711 = 1,030,865
(f) How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar,
be arranged in a row so that at least two candy bars of the same type are adjacent?
220 – 2 = 1,048,574
(g) How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar,
be arranged in a row so that at least two Snicker’s bars are adjacent?
220 – F(20) = 1,048,576 – 17,711 = 1,030,865
(h) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row?
20!
——— = 167,960
11! 9!
(i) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so
that candy bars of the same type are never adjacent?
0
(j) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so
that Milky Way bars are never adjacent?
0
(k) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so
that Snicker’s bars are never adjacent?
C(12,9) = 220
4.-continued
(l) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so
that the Milky Way bars are all adjacent?
10
(m) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so
that the Snicker’s bars are all adjacent?
12
(n) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so
that at least two candy bars of the same type are adjacent?
20!
——— – 0 = 167,960
11! 9!
(o) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so
that at least two Snicker’s bars are adjacent?
20!
——— – C(12,9) = 167,960 – 220 = 167,740
11! 9!
(p) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row
so that at least two Milky Way bars are adjacent?
20!
——— – 0 = 167,960
11! 9!
(q) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so
that candy bars of the same type are all adjacent?
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