1. L1L2L3n1n2n3 :With no restrictions, we have 26·26·26·10·10·10 = 17; 576; 000 permutations. With the restriction of no repeats,we have 26 · 25 · 24 · 10 · 9 · 8 = 11; 232; 000 permutations.L1L2n1n2L3L4:With no restrictions, we have 26·26·10·10·26·26 = 45; 697; 600 permutations. With the restriction of no repeats,we have 26 · 25 · 10 · 9 · 24 · 23 = 32; 292; 000 permutations.2. Call E the event all senators are male”. Then the probability of E happening isP (E) = jEjjSj =ways of sampling 3 from 80, no replacement and no orderways of sampling 3 from 100, no replacement and no order=80 3 100 3 =80·79·783·2100·99·983·2= 0:5083. There are many correct answers here.(a) One example is aRb , a ≤ b.aRa 8a 2 N [ f0g, since a ≤ a 8a 2 N [ f0g. Therefore, R is reflexive.8a; b; c 2 N [ f0g, if a ≤ b and b ≤ c, then a ≤ c. Therefore, R is transitive.1 ≤ 2, but 2 6≤ 1. Therefore, R is not symmetric.(b) One example is aRb , ab > 0.8a; b 2 N, if ab > 0 then ba > 0. Therefore, R is symmetric.8a; b; c 2 N, if ab > 0 and bc > 0, then a and b have the same sign and b and c have the same sign. Hence,a and c have the same sign. Therefore, ac > 0.0 · 0 6> 0. Therefore, R is not reflexive.4. This relation is symmetric and reflexive (prove it if you like), but not transitive. For instance, let x = 1, y = 2and z = 3. Then jx – yj = j1 – 2j = 1 ≤ 1;jy – zj = j2 – 3j = 1 ≤ 1;jx – zj = j1 – 3j = 2 6≤ 1:andbut Therefore, R is not an equivalence relation.5.• 2a – b = 2a – b 8(a; b) 2 R2. Therefore (a; b)R(a; b) 8(a; b) 2 R2 and R is reflexive.• 8(a; b); (c; d) 2 R2; if 2a-b = 2c-d, then 2c-d = 2a-b. Therefore, (a; b)R(c; d) ) (c; d)R(a; b) 8(a; b); (c; d) 2R2 and R is symmetric.• 8(a; b); (c; d); (e; f) 2 R2, if 2a – b = 2c – d and 2c – d = 2e – f, then 2a – b = 2e – f. Therefore,(a; b)R(c; d) ^ (c; d)R(e; f) ) (a; b)R(e; f) 8(a; b); (c; d); (e; f) 2 R2 and R is transitive.Therefore, R is an equivalence relation. There are many correct answers to the next part; for instance(1; 2); (2; 4) -1 3; -2 3 2 [(1; 2)].

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