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I subset). The second term counts the number 0 this choice if the two distinguished elements
_ ' " elements to be the rest of the subset). Note are the same (choose it, then choose any subse ' that this works even if n. = 1. 10. 3) 0(6 + 12 — 1 12) = C(17,12) = 6188 b) 0(6 + 36 — 1, 36) = C(41,36) = 749,398
' ' 6 = 12 croissants. This leaves one dozen left to as in part (3), namely C(6+12—1, 12) = C(17,12) = 6188. This can be done in C (6 +21 — 1, 21) = C(26, 21) = 65780 ways, since once we
' ' ' ' . Since there are C(6+24 — 1,24) = C(29,24) = 118755 there must be 118755 — 65780 = 52,975 ways to choose = 16 croissants without be C(6 + 15 — 1,15) = 090,15) = 15504 ways to make the the broccoli restriction by choosing at least four broccoli croissants, there would be C(G + 11 — 1,11) = C(16, 11) = 4368 choices. Therefore the number of ways to make the selection without violating the restriction ' .l. is 15504  4368 — 11,136.  > 2. This uses up 12 of the 29 total required, so the problem is the same as ﬁnding the number of solutions to 1”, + at; + mg + 3321+ as}, + 53% = 17 with each a nonnegative integer. The number of solutions is therefore 0 (6 + 17 ~— 1, 17) = C(22, 17) = 26,334. b) The restrictions use up 22 of the total, leaving a free total of 7. Therefore the answer is C (6 + 7 — 1, 7) :2 C(12,7) = 792. 
c) The number of solutions without restriction is C (6 + 29 — 1, 29) = C (34, 29) =2 278256. The number of solution violating the restriction by having 11:1 2 6 is C (6 + 23 — 1 23) = C (28, 23) = 98280. Therefore the answer is 278256 — 98280 = 179,976.
has required) but without the restriction on $1 is C (6 + 20  I d) The number of solutions with :52 2 9 (
1,20) = C (25, 20) := 53130. The number of solution violating the additional restriction by having :31 Z 8 is C(6 + 12  1,12) = C(17, 12) = 6188. Therefore the answer is 53130 — 6188 = 46,942. C(17,12) = 6188 ways. ' problem leaves us free to pick which boxes get which numbers of balls. There are several ways to count this. .Here is one. Line up the 15 objects in a row (15! ways to do that), and line up the ﬁve boxes in a row (5! ways to do that). Now put the ﬁrst object into the ﬁrst box, thenext two into the second 'nto the third box, and so on. This overcounts by a factor of 1!  2!  3!  4!  5!, since there
affecting the result. Therefore the answer is Without loss of generality assume that $6 = .
the number of nonnegative integer solutions to
each 3:, g 8 was moot, since the sum was only solutions. Therefore the answer is 6  70 z 420. 1:1 +332 +,:C3 l 334 +315 :2 4 (note t
4). By Theorem 2 there are 0(5 + 4 — 1,4) = C(8,4) = 70 sed from the usual roles, as, for example, in Theorem 2.) of these. Then it — (q1 !q2 +_  +q,) =
Therefore by Theorem 2, 28. (Note that the roles of the letters as and 7" here are rever We can choose the required objects ﬁrst, and there are q1 + «32 + + q,
‘ remain to be chosen. There are still 7' types. "9r)‘1a(n‘91Q2—”'_qr))= put two into one box and the So the answer is 3. 81' .____.__.._._._.— a 7.35 x 10101. (4!)4  (3!)11 r (2015
(The arithmetic was done with M ople.) integer solutions to m +n2+ * +nm = 71, number of ways to specify a term,
. Note that the coefﬁcients C for these terms can be computed using which by Theorem 2 is C (m + n — 1, ‘11)
Theorem 3——see Exercise 63. = CI(6,4) = 15 terms, and the coefﬁcients come 64. From Exercise 62, we know that there are C (3 + 4 ~— 1,4)
3 + 41332: + 4:333 + 4y3z + 492:3 + 61132312 + 6132,22 + from Exercise 63. The answer is :34 + y4 + Z4 + 4:173:24 + 4mg SECTION 5 .6 Generating Permutations and Combinations 2. 156423, 165432, 231456, 231465, 234561, 314562, 432561, 435612, 541236, 543216, 654312, 654321 6. The ﬁrst subset corresponds to the bit string 0000,
the bit string 0001., namely the set The next bit string is ,
0011 which corresponds to the set {3, 4}. We continue in this manner, 1 {2,3}... {213,4}, {1}, {1:4}, {173}: {1,314}: {1,2}: {112,4}: {132:3}, {152,314} ...
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 Spring '09

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