STATS 1L03 Lecture 4: 5.3 Counting Elements
27 Nov 2017
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How many ways can we come in the back and out the. (cid:1006) f(cid:396)o(cid:374)t of the hall? door (4 in) 6 (2out) = 8 routes through room! Chris has 5 hats, 6 shirts, ii pants. 4 4 doors-y e. g. son that one shirt. If i make a series of choices, with options my m2, my etc. (m, options i, my an choice 2 etc. ) Total number of combined options/ possibilities ie. m, o m 2 m3 etc. a shahs: 2shirt. 24+6 = 30 total outfits add parallel options: scrapes. Solution the sequential choices: (10 options). (12 options) (7)=840 possible teams. Solution: two parallel sequences (5 ford) (6 colours) +(3 toyota) (4 colours) 630 + 12=42) Now say i want to insure with one of 3 companies (3 insurance). (42 possible cars) (3 ins). (5 ford. 4 or. equivalently 355. 6+3. 4) results are the same regardless. Special cases of sequential choices: (cid:395) i a ((cid:272)a(cid:374) co(cid:396)e.
