Resoluções lista 01 IPE

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BC0406: Intr. à Prob. e à Estatística UFABC Resolução da Lista 01 v4 
 
 
Fernando Freitas Alves fernando.freitas@aluno.ufabc.edu.br 05/03/13 – pág. 1/8 
𝟏. 𝑃𝑎𝑟𝑎 𝑐𝑎𝑑𝑎 𝑝𝑜𝑟𝑡𝑎 𝑞𝑢𝑒 é 𝑝𝑜𝑠𝑠í𝑣𝑒𝑙 𝑒𝑛𝑡𝑟𝑎𝑟 (6), é 𝑝𝑜𝑠𝑠í𝑣𝑒𝑙 𝑠𝑎𝑖𝑟 𝑝𝑜𝑟 𝑞𝑢𝑎𝑙𝑞𝑢𝑒𝑟 𝑜𝑢𝑡𝑟𝑎 𝑝𝑜𝑟𝑡𝑎 (6): 
𝐸𝑛𝑡𝑟𝑎 𝑝𝑒𝑙𝑎 1ª: 𝑠𝑎𝑖 𝑝𝑒𝑙𝑎 1,2,3,4,5 𝑜𝑢 6ª 𝑝𝑜𝑟𝑡𝑎 → 6 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
𝐸𝑛𝑡𝑟𝑎 𝑝𝑒𝑙𝑎 2ª: 𝑠𝑎𝑖 𝑝𝑒𝑙𝑎 1,2,3,4,5 𝑜𝑢 6ª 𝑝𝑜𝑟𝑡𝑎 → 6 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
𝐸𝑛𝑡𝑟𝑎 𝑝𝑒𝑙𝑎 3ª: 𝑠𝑎𝑖 𝑝𝑒𝑙𝑎 1,2,3,4,5 𝑜𝑢 6ª 𝑝𝑜𝑟𝑡𝑎 → 6 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
⋮ 
𝐸𝑛𝑡𝑟𝑎 𝑝𝑒𝑙𝑎 6ª: 𝑠𝑎𝑖 𝑝𝑒𝑙𝑎 1,2,3,4,5 𝑜𝑢 6ª 𝑝𝑜𝑟𝑡𝑎 → 6 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
𝐿𝑜𝑔𝑜: 6 + 6 + 6 + 6 + 6 + 6 = 6 · 6 = 36 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑑𝑒 𝑒𝑛𝑡𝑟𝑎𝑑𝑎𝑠 
 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑑𝑒 𝑠𝑎í𝑑𝑎𝑠 
 
𝟐. 𝐴𝑔𝑜𝑟𝑎, 𝑝𝑎𝑟𝑎 𝑐𝑎𝑑𝑎 𝑝𝑜𝑟𝑡𝑎 𝑞𝑢𝑒 é 𝑝𝑜𝑠𝑠í𝑣𝑒𝑙 𝑒𝑛𝑡𝑟𝑎𝑟 (6), é 𝑝𝑜𝑠𝑠í𝑣𝑒𝑙 𝑠𝑎𝑖𝑟 𝑝𝑜𝑟 𝑎𝑝𝑒𝑛𝑎𝑠 𝑝𝑜𝑟𝑡𝑎𝑠 𝑑𝑖𝑠𝑖𝑛𝑡𝑎𝑠 (5): 
𝐸𝑛𝑡𝑟𝑎 𝑝𝑒𝑙𝑎 1ª: 𝑠𝑎𝑖 𝑝𝑒𝑙𝑎 2,3,4,5 𝑜𝑢 6ª 𝑝𝑜𝑟𝑡𝑎 → 5 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
𝐸𝑛𝑡𝑟𝑎 𝑝𝑒𝑙𝑎 2ª: 𝑠𝑎𝑖 𝑝𝑒𝑙𝑎 1,3,4,5 𝑜𝑢 6ª 𝑝𝑜𝑟𝑡𝑎 → 5 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
𝐸𝑛𝑡𝑟𝑎 𝑝𝑒𝑙𝑎 3ª: 𝑠𝑎𝑖 𝑝𝑒𝑙𝑎 1,2,4,5 𝑜𝑢 6ª 𝑝𝑜𝑟𝑡𝑎 → 5 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
⋮ 
𝐸𝑛𝑡𝑟𝑎 𝑝𝑒𝑙𝑎 6ª: 𝑠𝑎𝑖 𝑝𝑒𝑙𝑎 1,2,3,4 𝑜𝑢 5ª 𝑝𝑜𝑟𝑡𝑎 → 5 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
𝐿𝑜𝑔𝑜: 5 + 5 + 5 + 5 + 5 + 5 = 6 · 5 = 30 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑑𝑒 𝑒𝑛𝑡𝑟𝑎𝑑𝑎𝑠 
 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑑𝑒 𝑠𝑎í𝑑𝑎𝑠 
 
𝟑. 𝐸𝑛𝑡𝑟𝑒 10.000 𝑒 100.000 𝑡𝑒𝑚𝑜𝑠 𝑑𝑒 5 𝑎 6 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠 𝑑𝑒 𝑒𝑠𝑝𝑎ç𝑜 𝑑𝑖𝑠𝑝𝑜𝑛í𝑣𝑒𝑖𝑠: __ __ __ __ __ 𝑜𝑢 __ __ __ __ __ __ . 
𝑁𝑜 𝑒𝑛𝑡𝑎𝑛𝑡𝑜, 𝑜 ú𝑛𝑖𝑐𝑜 𝑛ú𝑚𝑒𝑟𝑜 𝑒𝑛𝑡𝑟𝑒 10.000 𝑒 100.000 𝑐𝑜𝑚 6 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠 é 𝑜 𝑛ú𝑚𝑒𝑟𝑜 100.000. 
𝐶𝑜𝑚𝑜 𝑑𝑒𝑣𝑒𝑚𝑜𝑠 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑟 𝑎𝑝𝑒𝑛𝑎𝑠 𝑜𝑠 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠 6,7 𝑜𝑢 8, 𝑜𝑠 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠 𝑑𝑒 100.000, 𝑞𝑢𝑒 𝑠ã𝑜 1 𝑒 0 𝑛ã𝑜 𝑝𝑜𝑑𝑒𝑚 
𝑎𝑝𝑎𝑟𝑒𝑐𝑒𝑟. 𝐴𝑠𝑠𝑖𝑚, 𝑡𝑜𝑑𝑎 𝑎 𝑛𝑜𝑠𝑠𝑎 𝑔𝑎𝑚𝑎 𝑑𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑡𝑒𝑟á 𝑎𝑝𝑒𝑛𝑎𝑠 𝟓 𝒄𝒂𝒓𝒂𝒄𝒕𝒆𝒓𝒆𝒔. 
𝑃𝑜𝑑𝑒𝑚𝑜𝑠 𝑢𝑡𝑙𝑖𝑙𝑖𝑧𝑎𝑟 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑐𝑜𝑚𝑜 67.678 𝑜𝑢 88.888. 𝐸𝑛𝑡ã𝑜, 𝑐𝑎𝑑𝑎 𝑒𝑠𝑝𝑎ç𝑜 𝑑𝑒 𝑝𝑜𝑠𝑖çã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝟑 𝒑𝒐𝒔𝒔𝒊𝒃𝒊𝒍𝒊𝒅𝒂𝒅𝒆𝒔. 
𝐿𝑜𝑔𝑜: 3 · 3 · 3 · 3 · 3 = 35 = 243 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑖𝑛𝑡𝑒𝑖𝑟𝑜𝑠 
 𝑒𝑠𝑝𝑎ç𝑜𝑠 𝑑𝑖𝑠𝑝𝑜𝑛í𝑣𝑒𝑖𝑠 (𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠) 
 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑒𝑚 𝑐𝑎𝑑𝑎 𝑒𝑠𝑝𝑎ç𝑜 (𝑎𝑙𝑓𝑎𝑏𝑒𝑡𝑜) 
 
BC0406: Intr. à Prob. e à Estatística UFABC Resolução da Lista 01 v4 
 
 
Fernando Freitas Alves fernando.freitas@aluno.ufabc.edu.br 05/03/13 – pág. 2/8 
 
𝟒. 𝐸𝑛𝑡𝑟𝑒 10.000 𝑒 100.000 𝑡𝑒𝑚𝑜𝑠 𝑑𝑒 5 𝑎 6 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠 𝑑𝑒 𝑒𝑠𝑝𝑎ç𝑜 𝑑𝑖𝑠𝑝𝑜𝑛í𝑣𝑒𝑖𝑠: __ __ __ __ __ 𝑜𝑢 __ __ __ __ __ __ . 
𝑁𝑜 𝑒𝑛𝑡𝑎𝑛𝑡𝑜, 𝑜 ú𝑛𝑖𝑐𝑜 𝑛ú𝑚𝑒𝑟𝑜 𝑒𝑛𝑡𝑟𝑒 10.000 𝑒 100.000 𝑐𝑜𝑚 6 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠 é 𝑜 𝑛ú𝑚𝑒𝑟𝑜 100.000. 
𝐶𝑜𝑚𝑜 𝑑𝑒𝑣𝑒𝑚𝑜𝑠 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑟 𝑎𝑝𝑒𝑛𝑎𝑠 𝑜𝑠 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠 0,6,7 𝑜𝑢 8, 𝑜 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒 0 𝑛ã𝑜 𝑝𝑜𝑑𝑒 𝑎𝑝𝑎𝑟𝑒𝑐𝑒𝑟 𝑛𝑜 𝑐𝑜𝑚𝑒ç𝑜. 
𝑇𝑜𝑑𝑎 𝑎 𝑛𝑜𝑠𝑠𝑎 𝑔𝑎𝑚𝑎 𝑑𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑡𝑒𝑟á 𝑎𝑝𝑒𝑛𝑎𝑠 𝟓 𝒄𝒂𝒓𝒂𝒄𝒕𝒆𝒓𝒆𝒔. 
𝑃𝑜𝑑𝑒𝑚𝑜𝑠 𝑢𝑡𝑙𝑖𝑙𝑖𝑧𝑎𝑟 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑐𝑜𝑚𝑜 61.786 𝑜𝑢 88.111. 𝐸𝑛𝑡ã𝑜, 𝑐𝑎𝑑𝑎 𝑒𝑠𝑝𝑎ç𝑜 𝑑𝑒 𝑝𝑜𝑠𝑖çã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝟒 𝒑𝒐𝒔𝒔𝒊𝒃𝒊𝒍𝒊𝒅𝒂𝒅𝒆𝒔. 
𝐿𝑜𝑔𝑜: 3 · 4 · 4 · 4 · 4 = 3 · 44 = 768 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑖𝑛𝑡𝑒𝑖𝑟𝑜𝑠 
 𝑒𝑠𝑝𝑎ç𝑜𝑠 𝑑𝑖𝑠𝑝𝑜𝑛í𝑣𝑒𝑖𝑠 (𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠) 𝑚𝑒𝑛𝑜𝑠 𝑜 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑜 𝑞𝑢𝑒 é 𝑟𝑒𝑠𝑡𝑟𝑖𝑡𝑜 
 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑒𝑚 𝑐𝑎𝑑𝑎 𝑒𝑠𝑝𝑎ç𝑜 (𝑎𝑙𝑓𝑎𝑏𝑒𝑡𝑜) 
 𝑎𝑝𝑒𝑛𝑎𝑠 6,7 𝑒 8 
 
𝟓. 𝐸𝑛𝑡𝑟𝑒 1.000 𝑒 9.999 (𝑖𝑛𝑐𝑙𝑢𝑠𝑖𝑣𝑒) 𝑡𝑒𝑚𝑜𝑠 𝟒 𝒄𝒂𝒓𝒂𝒄𝒕𝒆𝒓𝒆𝒔 𝑑𝑒 𝑒𝑠𝑝𝑎ç𝑜 𝑑𝑖𝑠𝑝𝑜𝑛í𝑣𝑒𝑖𝑠: __ __ __ __ __ . 
𝐶𝑜𝑚𝑜 𝑑𝑒𝑣𝑒𝑚𝑜𝑠 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑟 𝑎𝑝𝑒𝑛𝑎𝑠 𝑜𝑠 𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠 𝑑𝑖𝑠𝑡𝑖𝑛𝑡𝑜𝑠, 𝑑𝑒𝑝𝑜𝑖𝑠 𝑑𝑒 𝑠𝑒𝑙𝑒𝑐𝑖𝑜𝑛𝑎𝑟 𝑐𝑎𝑑𝑎 𝑑í𝑔𝑖𝑡𝑜, 𝑜 𝑝𝑟ó𝑥𝑖𝑚𝑜 𝑓𝑖𝑐𝑎 
𝑐𝑜𝑚 𝑜 𝑛ú𝑚𝑒𝑟𝑜 𝑑𝑒 𝒑𝒐𝒔𝒔𝒊𝒃𝒊𝒍𝒊𝒅𝒂𝒅𝒆𝒔 𝒍𝒊𝒎𝒊𝒕𝒂𝒅𝒐 𝑎 𝑛ã𝑜 𝑠𝑒𝑙𝑒𝑐𝑖𝑜𝑛𝑎𝑟 𝑛𝑒𝑛ℎ𝑢𝑚 𝑑𝑜𝑠 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟𝑒𝑠. 
𝑃𝑜𝑑𝑒𝑚𝑜𝑠 𝑢𝑡𝑙𝑖𝑙𝑖𝑧𝑎𝑟 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑐𝑜𝑚𝑜 61.786 𝑚𝑎𝑠 𝑛ã𝑜 𝑐𝑜𝑚𝑜 88.111. 
𝐿𝑜𝑔𝑜: 9 · 9 · 8 · 7 =
9 · 9!
(10 − 5)!
= 4.536 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑖𝑛𝑡𝑒𝑖𝑟𝑜𝑠 𝑑𝑖𝑠𝑡𝑖𝑛𝑡𝑜𝑠 
 𝑒𝑠𝑝𝑎ç𝑜𝑠 𝑑𝑖𝑠𝑝𝑜𝑛í𝑣𝑒𝑖𝑠 (𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠) 
 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑒𝑚 𝑐𝑎𝑑𝑎 𝑒𝑠𝑝𝑎ç𝑜 (𝑎𝑙𝑓𝑎𝑏𝑒𝑡𝑜) 
𝑃𝑎𝑟𝑎 𝑠𝑒𝑟 𝑝𝑎𝑟, 𝑜 𝑛ú𝑚𝑒𝑟𝑜 𝑝𝑟𝑒𝑐𝑖𝑠𝑎 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑎𝑝𝑒𝑛𝑎𝑠 𝑐𝑜𝑚 2,4,6,8 𝑜𝑢 0. 𝑁𝑜 𝑒𝑛𝑡𝑎𝑛𝑡𝑜, 𝑒𝑥𝑖𝑠𝑡𝑒𝑚 𝑒𝑣𝑒𝑛𝑡𝑜𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠: 
𝑄𝑢𝑎𝑛𝑑𝑜 𝑡𝑖𝑣𝑒𝑟 𝟒 í𝒎𝒑𝒂𝒓𝒆𝒔 𝑒 𝒏𝒆𝒏𝒉𝒖𝒎 𝒑𝒂𝒓 → 5 · 4 · 3 · 2 · 5 = 600 𝑝𝑎𝑟𝑒𝑠 
𝑄𝑢𝑎𝑛𝑑𝑜 𝑡𝑖𝑣𝑒𝑟 𝟑 í𝒎𝒑𝒂𝒓𝒆𝒔 𝑒 𝟏 𝒑𝒂𝒓 → 5 · 4 · 3 · 5 · 4 = 1200 𝑝𝑎𝑟𝑒𝑠 
𝑄𝑢𝑎𝑛𝑑𝑜 𝑡𝑖𝑣𝑒𝑟 𝟐 í𝒎𝒑𝒂𝒓𝒆𝒔 𝑒 𝟐 𝒑𝒂𝒓𝒆𝒔 → 5 · 4 · 5 · 4 · 3 = 1200 𝑝𝑎𝑟𝑒𝑠 
𝑄𝑢𝑎𝑛𝑑𝑜 𝑡𝑖𝑣𝑒𝑟 𝟏 í𝒎𝒑𝒂𝒓 𝑒 𝟑 𝒑𝒂𝒓𝒆𝒔 → 5 · 5 · 4 · 3 · 2 = 600 𝑝𝑎𝑟𝑒𝑠 
𝑄𝑢𝑎𝑛𝑑𝑜 𝑡𝑖𝑣𝑒𝑟 𝒏𝒆𝒏𝒉𝒖𝒎 í𝒎𝒑𝒂𝒓 𝑒 𝟒 𝒑𝒂𝒓𝒆𝒔 → 4 · 4 · 3 · 2 · 1 = 5 · 4 · 3 · 2 · 1 − 4 · 3 · 2 · 1 = 96 𝑝𝑎𝑟𝑒𝑠 
𝑅𝑒𝑝𝑎𝑟𝑒 𝑞𝑢𝑒 𝑜 ú𝑙𝑡𝑖𝑚𝑜 𝑜𝑏𝑟𝑖𝑔𝑎𝑡𝑜𝑟𝑖𝑎𝑚𝑒𝑛𝑡𝑒 é 𝑝𝑎𝑟. 
𝐿𝑜𝑔𝑜: 600 + 1200 + 1200 + 600 + 120 − 24 = ∑
5!
𝑖!
·
5!
(5 − 𝑖)!
5
𝑖=1
− 4! = 3.696 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑝𝑎𝑟𝑒𝑠 𝑑𝑖𝑠𝑡𝑖𝑛𝑡𝑜𝑠 
 𝑛ú𝑚𝑒𝑟𝑜 𝑑𝑒 𝑝𝑎𝑟𝑒𝑠 𝑝𝑜𝑠𝑠í𝑣𝑒𝑖𝑠 
 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑑𝑒 í𝑚𝑝𝑎𝑟𝑒𝑠 𝑝𝑜𝑠𝑠í𝑣𝑒𝑖𝑠 
 𝑒𝑠𝑝𝑎ç𝑜𝑠 𝑑𝑖𝑠𝑝𝑜𝑛í𝑣𝑒𝑖𝑠 (𝑐𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑒𝑠) 
 
BC0406: Intr. à Prob. e à Estatística UFABC Resolução da Lista 01 v4 
 
 
Fernando Freitas Alves fernando.freitas@aluno.ufabc.edu.br 05/03/13 – pág. 3/8 
 
𝟔. 𝒂) 
12!
10!
=
12 · 11 · 10!
10!
= 12 · 11 = 12 · (10 + 1) = 120 + 12 = 132 
𝒃) 
𝑛!
(𝑛 − 𝑟)!
=
𝑛 · (𝑛 − 1) · (𝑛 − 2) · ⋯ · (𝑛 − 𝑟 + 1) · (𝑛 − 𝑟)!
(𝑛 − 𝑟)!
= 𝑛 · (𝑛 − 1) · (𝑛 − 2) · ⋯ · (𝑛 − 𝑟 + 1) 
 
𝟕. 𝒂) 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑛𝑑𝑜 𝑞𝑢𝑒 𝑠ó 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑎𝑛𝑑𝑎𝑟 𝑝𝑎𝑟𝑎 𝑜 𝑛𝑜𝑟𝑡𝑒 (𝑁) 𝑒 𝑝𝑎𝑟𝑎 𝑜 𝑙𝑒𝑠𝑡𝑒 (𝐿) 𝑒 𝑐𝑜𝑚𝑜 𝐴 𝑒𝑠𝑡á 𝑎 10 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 
𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝑑𝑒 𝐶, 𝑜 𝑛ú𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑚𝑖𝑛ℎ𝑜𝑠 𝑝𝑜𝑠𝑠í𝑣𝑒𝑖𝑠 é 𝑢𝑚𝑎 𝑝𝑒𝑟𝑚𝑢𝑡𝑎çã𝑜 𝑑𝑒 4 𝑝𝑎𝑟𝑎 𝑁 𝑒 6 𝑝𝑎𝑟𝑎 𝐿: 
𝐸𝑥𝑒𝑚𝑝𝑙𝑜𝑠: 𝐿 𝐿 𝐿 𝐿 𝑁 𝑁 𝑁 𝑁 𝑁 𝑁, 𝐿 𝑁 𝐿 𝐿 𝐿 𝑁 𝑁 𝑁 𝑁 𝑁, 𝐿 𝑁 𝑁 𝐿 𝑁 𝑁 𝐿 𝑁 𝑁 𝐿. 
𝐿𝑜𝑔𝑜: (
10
4, 6
) = (
10
4
) = (
10
6
) =
10!
4! 6!
= 5 · 3 · 2 · 7 = 210 𝑓𝑜𝑟𝑚𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒃) (
4
2, 2
) · (
6
2, 4
) = (
4
2
) · (
6
2
) =
4!
2! 2!
+
6!
2! 4!
= 3! + 3 · 5 = 21 𝑓𝑜𝑟𝑚𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒄) (
10
4, 6
) − (
4
2, 2
) = (
10
4
) − (
4
2
) =
10!
4! 6!
−
4!
2! 2!
= 10 · 7 · 3 − 3! = 204 𝑓𝑜𝑟𝑚𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒅) (
10
4, 6
) + (
6
2, 4
) = (
10
4
) + (
6
2
) =
10!
4! 6!
+
6!
2! 4!
= 10 · 7 · 3 + 3 · 5 = 225 𝑓𝑜𝑟𝑚𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝟖. (
20
2
) =
20!
2! (20 − 2!)
=
20 · 19 · 18!
2! 18!
=
20 · 19
2
= 10 · 19 = 190 𝑟𝑒𝑡𝑎𝑠 𝑙𝑖𝑔𝑎𝑛𝑑𝑜 𝑑𝑜𝑖𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 
(
20
3
) =
20!
3! (20 − 3!)
=
20 · 19 · 18 · 17!
3! 17!
=
20 · 19 · 18
3 · 2
= 10 · 19 · 6 = 1140 𝑟𝑒𝑡𝑎𝑠 𝑡𝑟𝑖â𝑛𝑔𝑢𝑙𝑜𝑠 𝑝𝑜𝑟 𝑡𝑟𝑖𝑝𝑙𝑎𝑠 
 
𝟗. 𝒂) 13! = 6.227.020.800 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒃) 6! 3! 4! = 103.680 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒄) (6! 3! 4!) · 3! = 622.080 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝟏𝟎. 𝒂) 
13!
3!
= 1.037.836.800 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒃) 
6!
3!
3! 4! = 17.280 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒄) (
6!
3!
3! 4!) 3! = 103.680 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝟏𝟏. 26𝟐 · 104 = (26 · 26) · (10 · 10 · 10 · 10) = 6.760.000 𝑝𝑙𝑎𝑐𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
26!
(26 − 2)!
·
10!
(10 − 4)!
= (26 · 25) · (10 · 9 · 8 · 7) = 3.276.000 𝑝𝑙𝑎𝑐𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
BC0406: Intr. à Prob. e à Estatística UFABC Resolução da Lista 01 v4 
 
 
Fernando Freitas Alves fernando.freitas@aluno.ufabc.edu.br 05/03/13 – pág. 4/8 
 
𝟏𝟐. (
26
3
) − [26 − (2 − 1)] (
26
3 − 2
) = (
26
3
) − 25 (
26
1
) = 1950 𝑐𝑜𝑛𝑗𝑢𝑛𝑡𝑜𝑠 𝑑𝑒 𝑙𝑒𝑡𝑟𝑎𝑠 𝑠𝑒𝑚 𝑝𝑎𝑟𝑒𝑠 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑜𝑠 
 
𝟏𝟑. 𝒂) (
7 + 6 + 4
3
) = (
17
3
) = 680 𝑒𝑠𝑐𝑜𝑙ℎ𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒃) (
7
3
) + (
6
3
) + (
4
3
) =
7 · 6 · 5 + 6 · 5 · 4 + 4 · 3 · 2
3 · 2
= 7 · 5 + 5 · 4 + 4 = 59 𝑒𝑠𝑐𝑜𝑙ℎ𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒄) (
7
2
) [(
6
1
) + (
4
1
)] + (
6
2
) [(
7
1
) + (
4
1
)] + (
4
2
) [(
7
1
) + (
6
1
)] = 
= (
7
2
) 10 + (
6
2
) 11 + (
4
2
) 13 =
7 · 6 · 10 + 6 · 5 · 11 + 4 · 3 · 13
2
 
= 7 · 6 · 5 + 3 · 5 · 11 + 2 · 3 · 13 = 453 𝑒𝑠𝑐𝑜𝑙ℎ𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝟏𝟒. 𝒂) 𝑀𝐼𝑆𝑆𝐼𝑆𝑆𝐼𝑃𝑃𝐼 → 11 𝑝𝑜𝑠𝑖çõ𝑒𝑠 / 𝐼, 𝑆 𝑟𝑒𝑝𝑒𝑡𝑒𝑚 4𝑥 / 𝑃 𝑟𝑒𝑝𝑒𝑡𝑒 2𝑥 
→ (
11
4,4,2
) =
11!
4! 4! 2!
= 11 · 9 · 7 · 5 · 5 · 2 = 34.650 𝑎𝑛𝑎𝑔𝑟𝑎𝑚𝑎𝑠 
𝒃) 𝐿𝐼𝑆𝑇𝐴 → 5 𝑝𝑜𝑠𝑖çõ𝑒𝑠 / 𝑛𝑒𝑛ℎ𝑢𝑚 𝑟𝑒𝑝𝑒𝑡𝑒 →
5!
0!
= 5! = 120 𝑎𝑛𝑎𝑔𝑟𝑎𝑚𝑎𝑠 
𝒄) 𝑃𝑅𝑂𝐵𝐴𝐵𝐼𝐿𝐼𝐷𝐴𝐷𝐸 → 14 𝑝𝑜𝑠𝑖çõ𝑒𝑠 𝑅 𝑟𝑒𝑝𝑒𝑡𝑒 1𝑥 𝑒 𝐵, 𝐴, 𝐼, 𝐷 𝑟𝑒𝑝𝑒𝑡𝑒𝑚 2𝑥 
→ (
13
1,2,2,2,2
) =
13!
1! 2! 2! 2! 2!
= 389.188.800 𝑎𝑛𝑎𝑔𝑟𝑎𝑚𝑎𝑠 
𝒅) 𝐵𝐴𝑁𝐴𝑁𝐴 → 6 𝑝𝑜𝑠𝑖çõ𝑒𝑠 / 𝐴 𝑟𝑒𝑝𝑒𝑡𝑒 3𝑥 / 𝑁 𝑟𝑒𝑝𝑒𝑡𝑒 2𝑥 → (
6
3,2
) =
6!
3! 2!
= 5 · 4 · 3 = 60 𝑎𝑛𝑎𝑔𝑟𝑎𝑚𝑎𝑠 
 
𝟏𝟓. (
5
2
) = 20 𝑎𝑝𝑒𝑟𝑡𝑜𝑠 𝑑𝑒 𝑚ã𝑜 
 
𝟏𝟔. (
3
2
) · 3 + 3 · 2 = 15 𝑏𝑒𝑖𝑗𝑜𝑠 
 
𝟏𝟕. (
23 − 1
4 − 1
) = 1540 𝑠𝑜𝑙𝑢çõ𝑒𝑠 𝑖𝑛𝑡𝑒𝑖𝑟𝑎𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑎𝑠 
 
𝟏𝟖. 𝑥1 + 𝑥2 = 7 → • _ • _ • _ • _ • _ • _ • (𝑐𝑎𝑑𝑎 𝑒𝑠𝑝𝑎ç𝑜 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎 𝑢𝑚𝑎 𝑑𝑖𝑣𝑖𝑠ã𝑜) 
𝐸𝑥𝑒𝑚𝑝𝑙𝑜: • • • | • • • • 𝑖𝑛𝑑𝑖𝑐𝑎 3 + 4 
𝐸𝑛𝑡ã𝑜: (
6
1
) = 6 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 𝑝𝑎𝑟𝑎 62 𝑡𝑜𝑡𝑎𝑖𝑠 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒𝑠 
BC0406: Intr. à Prob. e à Estatística UFABC Resolução da Lista 01 v4 
 
 
Fernando Freitas Alves fernando.freitas@aluno.ufabc.edu.br 05/03/13 – pág. 5/8 
𝐿𝑜𝑔𝑜: 
(6
1
)
62
=
6
36
=
1
6
 𝑑𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒 ≅ 17% 
 
𝟏𝟗. 𝒂) • • • ⋯ • | • • • ⋯ • | | | ⋯ | = (
𝑟 − 𝑘 + 𝑛 − 1
𝑟 − 𝑘
) = (
𝑟 − 𝑘 + 𝑛 − 1
𝑛 − 1
) 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 𝑘 𝑏𝑜𝑙𝑎𝑠 𝑛𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑐𝑎𝑖𝑥𝑎 
 1 𝑑𝑖𝑣𝑖𝑠ã𝑜 𝑝𝑎𝑟𝑎 𝑖𝑛𝑑𝑖𝑐𝑎𝑟 𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑐𝑎𝑖𝑥𝑎 
 𝑝𝑒𝑟𝑚𝑢𝑡𝑎çõ𝑒𝑠 𝑒𝑛𝑡𝑟𝑒: (𝒓 − 𝒌) 𝒃𝒐𝒍𝒂𝒔 𝒓𝒆𝒔𝒕𝒂𝒏𝒕𝒆𝒔 + (𝒏 − 𝟏) 𝒅𝒊𝒗𝒊𝒔õ𝒆𝒔 𝒅𝒆 𝒄𝒂𝒊𝒙𝒂𝒔 
𝑥1
′ + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛 = 𝑟 ; 𝑥𝑖 ∈ ℕ 
𝑥1
′ = 𝑥1 + 𝑘 
𝑘 + 𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛 = 𝑟 − 𝑘 ; 𝑥𝑖 ∈ ℕ 
𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛 = 𝑟 − 𝑘 ; 𝑥𝑖 ∈ ℕ 
∑ 𝑥𝑖
𝑛
𝑖=1
= 𝑟 − 𝑘 ; 𝑥𝑖 ∈ ℕ 
 
𝒃) 𝑥1
′ + 𝑥2
′ + 𝑥3
′ + ⋯ + 𝑥𝑛
′ = 𝑟 ; 𝑥𝑖
′ ∈ ℕ 
𝑥𝑖
′ = 𝑥𝑖 + 2 
(𝑥1 + 2) + (𝑥2 + 2) + (𝑥3 + 2) + ⋯ + (𝑥𝑛 + 2) = 𝑟 ; 𝑥𝑖 ∈ ℕ 
𝑛 · 2 + 𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛 = 𝑟 ; 𝑥𝑖 ∈ ℕ 
𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛 = 𝑟 − 2𝑛 ; 𝑥𝑖 ∈ ℕ 
∑ 𝑥𝑖
𝑛
𝑖=1
= 𝑟 − 2𝑛 ; 𝑥𝑖 ∈ ℕ 
 
 
𝒄) • • ⋯ • | • • • ⋯ • | | | ⋯ | = (𝑝 + 1) (
𝑟 − 𝑝 + 𝑛 − 3
𝑟 − 𝑝
) = (𝑝 + 1) (
𝑟 − 𝑝 + 𝑛 − 3
𝑛 − 3
) 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠 𝑑𝑖𝑓. 
 𝑠𝑒𝑙𝑒𝑐𝑖𝑜𝑛𝑎𝑟 𝑒𝑛𝑡𝑟𝑒 𝑎𝑠 (𝒑 + 𝟏) 𝒑𝒐𝒔𝒊çõ𝒆𝒔, 𝑎 𝑑𝑖𝑣𝑖𝑠ã𝑜 𝑑𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑐𝑎𝑖𝑥𝑎 
 1 𝑑𝑖𝑣𝑖𝑠ã𝑜 𝑝𝑎𝑟𝑎 𝑖𝑛𝑑𝑖𝑐𝑎𝑟 𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑐𝑎𝑖𝑥𝑎 
 𝑝𝑒𝑟𝑚𝑢𝑡𝑎çõ𝑒𝑠: (𝒓 − 𝒑) 𝒃𝒐𝒍𝒂𝒔 𝒓𝒆𝒔𝒕𝒂𝒏𝒕𝒆𝒔 + (𝒏 − 𝟐) − 𝟏 𝒅𝒊𝒗𝒊𝒔õ𝒆𝒔 𝒓𝒆𝒔𝒕𝒂𝒏𝒕. 
(𝑥1 + 𝑥2) + 𝑥3 + ⋯ + 𝑥𝑛 = 𝑟 ; 𝑥𝑖 ∈ ℕ 
𝑥1 + 𝑥2 = 𝑝 
𝑝 + 𝑥3 + ⋯ + 𝑥𝑛 = 𝑟 ; 𝑥𝑖 ∈ ℕ 
𝑥3 + ⋯ + 𝑥𝑛 = 𝑟 − 𝑝 ; 𝑥𝑖 ∈ ℕ 
EXTRA 
EXTRA 
BC0406: Intr. à Prob. e à Estatística UFABC Resolução da Lista 01 v4 
 
 
Fernando Freitas Alves fernando.freitas@aluno.ufabc.edu.br 05/03/13 – pág. 6/8 
∑ 𝑥𝑖
𝑛
𝑖=3
= 𝑟 − 𝑝 ; 𝑥𝑖 ∈ ℕ 
 
𝟐𝟎. 𝒂) • • • • • • | ||⋯ | = (
6 + 31 − 1
6
) = (
6 + 31 − 1
31 − 1
) = 1.947.792 𝑠𝑒𝑙𝑒çõ𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 𝑝𝑒𝑟𝑚𝑢𝑡𝑎çõ𝑒𝑠: 6 𝑏𝑜𝑙𝑎𝑠 + (31 − 1) 𝑑𝑖𝑣𝑖𝑠õ𝑒𝑠 𝑝𝑎𝑟𝑎 𝑠𝑎𝑏𝑜𝑟𝑒𝑠 
∑ 𝑥𝑖
31
𝑖=1
= 6 ; 𝑥𝑖 ∈ ℕ 
 
𝒃) 𝑥1
′ + 𝑥2
′ + 𝑥3
′ = 5 
𝑥1
′ = 5 − 𝑥1 
𝑥2
′ = 4 − 𝑥2 
𝑥3
′ = 2 − 𝑥3 
(5 − 𝑥1) + (4 − 𝑥2) + (2 − 𝑥3) = 5 
𝑥1 + 𝑥2 + 𝑥3 = (5 + 4 + 2) − 5 
• • • • • • | | = (
(5 + 4 + 2) − 5
3 − 1
) = (
6
2
) = 15 𝑠𝑒𝑙𝑒çõ𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝒄) 𝑥1
′ + 𝑥2
′ + 𝑥3
′ + 𝑥4
′ = 12 
𝑥𝑖
′ = 𝑥𝑖 + 2 
(𝑥1 + 2) + (𝑥2 + 2) + (𝑥3 + 2) + (𝑥4 + 2) = 12 
4 · 2 + 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 12 
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 12 − 4 · 2 
• • • • | | | = (
(12 − 4 · 2) + 4 − 1
4 − 1
) = (
7
3
) = 35 𝑠𝑒𝑙𝑒çõ𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝒅) 𝑥1
′ + 𝑥2
′ + 𝑥3
′ + 𝑥4
′ = 20 
𝑥𝑖
′ = 2𝑥𝑖
′′ ⇒ 2𝑥1
′′ + 2𝑥2
′′ + 2𝑥3
′′ + 2𝑥4
′′ = 20 
𝑥1
′′ + 𝑥2
′′ + 𝑥3
′′ + 𝑥4
′′ = 10 
𝑥𝑖
′′ = 8/2 − 𝑥𝑖 
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 4 · 8/2 − 10 
•
•
 
•
•
 
•
•
 
•
•
 
•
•
 | | | = (
4 · 8/2 − 10
4 − 1
) = (
6
3
) = 20 𝑠𝑒𝑙𝑒çõ𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
BC0406: Intr. à Prob. e à Estatística UFABC Resolução da Lista 01 v4 
 
 
Fernando Freitas Alves fernando.freitas@aluno.ufabc.edu.br 05/03/13 – pág. 7/8 
 
𝟐𝟏. 8𝑥8 → 64 𝑝𝑜𝑠𝑖çõ𝑒𝑠 𝑝𝑜𝑠𝑠í𝑣𝑒𝑖𝑠 → (
64
8
) (
64 − 8
8
) =
64!
8! (64 − 8)!
·
(64 − 8)!
8! (64 − 8 − 8)!
=
64!
8! 8! 48!
= 6.287.341.680.214.194.600 𝑝𝑜𝑠𝑖çõ𝑒𝑠 𝑝𝑜𝑠𝑠í𝑣𝑒𝑖𝑠 
𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎 𝑑𝑒 180° → 𝑐𝑎𝑑𝑎 𝑒𝑠𝑐𝑜𝑙ℎ𝑎 𝑖𝑚𝑝𝑙𝑖𝑐𝑎 𝑒𝑚 𝑢𝑚𝑎 𝑒𝑠𝑐𝑜𝑙ℎ𝑎 𝑑𝑜 𝑜𝑢𝑡𝑟𝑜 𝑙𝑎𝑑𝑜 
→ 64 · 1 · 62 · 1 · 60 · 1 · 58 · 1 · 56 · 1 · 54 · 1 · 52 · 1 · 50 · 1 = 108.569.051.136.000 𝑝𝑜𝑠𝑖çõ𝑒𝑠 𝑝𝑜𝑠𝑠í𝑣𝑒𝑖𝑠 
 
𝟐𝟐.
(22
11
)
𝟐!
= 352.716 𝑑𝑖𝑣𝑖𝑠õ𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝟐𝟑. 36525 −
365!
(365 − 25)!
= 6.489.401.033.174.836.353.736.053.814.447.511.804.192.464.972.003.836.221.236.578.125 𝑚𝑎𝑛𝑒𝑖𝑟𝑎𝑠 𝑑𝑖𝑓. 
 
𝟐𝟒. 
100 − (5 − 1)
(100
5
)
=
1
784245
≅ 0,0001275% 
 
𝟐𝟓. (𝑐1 + 1) + (𝑐2 + 1) + (𝑐3 + 1) + (𝑐4 + 1) = 18 ; 𝑐𝑖 ∈ ℕ, 1 ≤ 𝑖 ≤ 4 
𝑐1 + 𝑐2 + 𝑐3 + 𝑐4 = 18 − 4 · 1 
(
18 − 4 · 1 + 4 − 1
4 − 1
) = (
17
3
) = 680 𝑚𝑜𝑑𝑜𝑠 𝑑𝑒 𝑎𝑝𝑜𝑠𝑡𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝟐𝟔. 𝒂) (
8
5
) − (
2
2
) (
8 − 2
5 − 2
) = (
8
5
) − (
5
3
) = 46 𝑒𝑠𝑐𝑜𝑙ℎ𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒃) (
2
2
) (
8 − 2
5 − 2
) = (
5
3
) = 10 𝑒𝑠𝑐𝑜𝑙ℎ𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
 
𝟐𝟕. 𝒂) 𝑥1 + 𝑥2 + ⋯ + 𝑥𝑟 = 𝑛 ; 𝑥𝑖 ∈ ℕ, 1 ≤ 𝑖 ≤ 𝑟 
𝑆𝑜𝑙𝑢çõ𝑒𝑠 𝑛ã𝑜 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑎𝑠 → 𝑥𝑖 ≥ 0 → 𝑥𝑖 = 𝑥𝑖
′ − 1; 𝑥𝑖
′ > 0 
(𝑥1
′ − 1) + (𝑥2
′ − 1) + ⋯ + (𝑥𝑟
′ − 1) = 𝑛 ; 𝑥𝑖
′ ∈ ℕ∗ , 1 ≤ 𝑖 ≤ 𝑟 
𝑥1
′ + 𝑥2
′ + ⋯ + 𝑥𝑟
′ − 𝑟 =𝑛 ; 𝑥𝑖
′ ∈ ℕ∗, 1 ≤ 𝑖 ≤ 𝑟 
𝑥1
′ + 𝑥2
′ + ⋯ + 𝑥𝑟
′ = 𝑛 + 𝑟 ; 𝑥𝑖
′ ∈ ℕ∗, 1 ≤ 𝑖 ≤ 𝑟 
• • ⋯ • • | | | ⋯ | = (
𝑛 + 𝑟 − 1
𝑟 − 1
) ∎ 
𝒃) (
23 + 4 − 1
4 − 1
) = (
26
3
) = 2600 𝑠𝑜𝑙𝑢çõ𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
BC0406: Intr. à Prob. e à Estatística UFABC Resolução da Lista 01 v4 
 
 
Fernando Freitas Alves fernando.freitas@aluno.ufabc.edu.br 05/03/13 – pág. 8/8 
 
𝟐𝟖. 𝒂) (𝑥1 + 2) + (𝑥2 + 2) + (𝑥3 + 3) + (𝑥4 + 4) = 20 ; 𝑥𝑖 ∈ ℕ, 1 ≤ 𝑖 ≤ 4 
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 20 − (2 + 2 + 3 + 4) ; 𝑥𝑖 ∈ ℕ, 1 ≤ 𝑖 ≤ 4 
(
20 − (2 + 2 + 3 + 4) + 4 − 1
4 − 1
) = (
9
3
) = 84 𝑒𝑠𝑡𝑟𝑎𝑡é𝑔𝑖𝑎𝑠 𝑑𝑒 𝑎𝑝𝑙𝑖𝑐𝑎çã𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 
𝒃) 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 9 ; 𝑥𝑘 = 0 𝑒 𝑥𝑖 ∈ ℕ
∗ , 1 ≤ 𝑖 ≤ 3, 𝑖 ≠ 𝑘 
→ 4 (
9 − 1
3 − 1
) = 112 𝑒𝑠𝑡𝑟𝑎𝑡é𝑔𝑖𝑎𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜 𝑢𝑚𝑎 𝑏𝑜𝑙𝑠𝑎 𝑛ã𝑜 𝑠𝑒𝑗𝑎 𝑎𝑝𝑙𝑖𝑐𝑎𝑑𝑎 
 
𝟐𝟗. (
26
4
) − [26 − (2 − 1)] (
26
4 − 2
) = (
26
4
) − 25 (
26
2
) = 6825 𝑐𝑜𝑛𝑗𝑢𝑛𝑡𝑜𝑠 𝑑𝑒 𝑙𝑒𝑡𝑟𝑎𝑠 𝑠𝑒𝑚 𝑝𝑎𝑟𝑒𝑠 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑜𝑠