35 permutations of 3 things | My Assignment Tutor

1. There are 35 permutations of 3 things taken 5 at a time. But 35 = 0 since k > n.There are 53 permutations of 5 things taken 3 as a time, and 53 = (5-5!3)! = 60.2. There are 25 = 32 binary words of length 5. There is 1 word that does not contain a 1 (00000) and 1 wordthat does not contain a 0 (11111). So there are 32 – 2 = 30 words that contain at least one 0 and at leastone 1.3. There are 3 ways to assign a: f(a) = 1; f(a) = 2 and f(a) = 3. Then for each of those there are three waysto assign b, etc. So in total there are 3 · 3 · 3 · 3 = 81 different functions.4. 8 5 = 5!8! ×3! = 56.5. (i) 10 2 = 8!10! ×2! = 45 chords (try it!).(ii) 10 3 = 7!10! ×3! = 120 triangles.(iii) 10 6 = 4!10! ×6! = 210 hexagons.6. Recall that if jAj = n, then A has 2n subsets. To be proper and nonempty, we exclude A and ? and solve2n – 2 > 100: n ≥ 7.7. Recall Pn k=0 nk = 2n. We want 8 1 + 8 2 + : : : + 8 8, which equals 28 – 8 0 = 255.8. The Binomial Theorem with x = -1; y = 5; n = 13 shows that P13 r=0(-1)r13 r 513-r = (5 + (-1))13 = 413:9. (a) P(all faulty) = (0:12)3 ≈ 0:0017.(b) P(no fault) = (0:88)3 ≈ 0:6815(c) P(1 faulty) = 3 1 × (0:12 × 0:88 × 0:88) ≈ 0:278810. P (A [ B) = P (A) + P (B) – P (A B) = 0:70 + 0:60 – 0:35 = 0:95. You can also solve by Venn diagram;you’re given P (A B) = 0:35, P (A) = 0:7 and P (B) = 0:6.

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