Answer to Question #323672 in Discrete Mathematics for gift

Question #323672



1, In how many ways each 5 boys and 3 girls be seated around a table if:





i) There is no restriction?





ii, Boy Bi and girl Gi are not adjacent?





ji, No girls are adjacent?






2. Each user on a computer system has a password, which is six to eight characters





long. where each character is an upper case letter or a digit Fach password must





contain at least one digit. How many possible passwords are there?






3.how many ways are there to distribute 10 identical bones to 4 dogs if each dog must





stat least | bone and Fifi may not receive more than 3 bones’?






1
Expert's answer
2022-04-07T08:23:28-0400

"i:\\\\8!=40320\\\\ii:\\\\Find\\,\\,the\\,\\,variants\\,\\,when\\,\\,Bi\\,\\,and\\,\\,Gi\\,\\,are\\,\\,adjacent:\\\\8 variants\\,\\,to\\,\\,choose\\,\\,a\\,\\,pair\\,\\,of\\,\\,seats\\,\\,for\\,\\,them\\,\\,\\left( we\\,\\,choose\\,\\,the\\,\\,chair\\,\\,which\\,\\,is\\,\\,to\\,\\,the\\,\\,left\\,\\,if\\,\\,walking\\,\\,clockwise \\right) ,\\\\2 variants\\,\\,to\\,\\,seat\\,\\,them,\\\\6!=720\\,\\,variants\\,\\,to\\,\\,seat\\,\\,other\\,\\,kids.That\\,\\,is\\,\\,\\\\8\\cdot 2\\cdot 720=11520\\\\variants.\\\\Then\\,\\,the\\,\\,answer\\,\\,is\\\\40320-11520=28800\\\\variants.\\\\ji:\\\\The\\,\\,possible\\,\\,chairs\\,\\,for\\,\\,girls\\,\\,are\\\\1,3,5\\\\1,3,6\\\\1,3,7\\\\1,4,6\\\\1,4,7\\\\1,5,7\\\\2,4,6\\\\2,4,7\\\\2,4,8\\\\2,5,7\\\\2,5,8\\\\2,6,8\\\\3,5,7\\\\3,5,8\\\\3,6,8\\\\4,6,8\\\\Thus\\,\\,there\\,\\,are\\,\\,16ways\\,\\,to\\,\\,seat\\,\\,the\\,\\,girls.\\\\3!=6\\,\\,variants\\,\\,of\\,\\,seating\\,\\,girls.\\\\5!=120\\,\\,variants\\,\\,of\\,\\,seating\\,\\,boys.\\\\Total\\,\\,is\\\\16\\cdot 6\\cdot 120=11520"


"26 letters\\\\10 digits\\\\2:\\\\Passwords\\,\\,of\\,\\,6 digits: \\\\P_{36}^{6}-P_{26}^{6}=\\frac{36!}{30!}-\\frac{26!}{20!}\\\\Passwords\\,\\,of\\,\\,7 digits: \\\\P_{36}^{7}-P_{26}^{7}=\\frac{36!}{29!}-\\frac{26!}{19!}\\\\Passwords\\,\\,of\\,\\,8 digits: \\\\P_{36}^{8}-P_{26}^{8}=\\frac{36!}{28!}-\\frac{26!}{18!}\\\\Total\\\\\\frac{36!}{30!}-\\frac{26!}{20!}+\\frac{36!}{29!}-\\frac{26!}{19!}+\\frac{36!}{28!}-\\frac{26!}{18!}=1.1971\\times 10^{12}"


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