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At-examination-a-student-most-answer-to-8-question-betwen-a-total-of-10-question-1-How-many-many-choices-is-possible-2-How-many-choices-is-possible-if-a-student-most-answer-to-3-questions-at-leas




Question Number 102243 by mathocean1 last updated on 07/Jul/20
At examination, a student most  answer to 8 question betwen a total  of 10 question.  1) How many many choices is   possible?  2)How many choices is possible if a  student most answer to 3 questions  at least?  3)How many choices is possible if a  student most answer to at least 4  questions of  the five first questions?
$${At}\:{examination},\:{a}\:{student}\:{most} \\ $$$${answer}\:{to}\:\mathrm{8}\:{question}\:{betwen}\:{a}\:{total} \\ $$$${of}\:\mathrm{10}\:{question}. \\ $$$$\left.\mathrm{1}\right)\:{How}\:{many}\:{many}\:{choices}\:{is}\: \\ $$$${possible}? \\ $$$$\left.\mathrm{2}\right){How}\:{many}\:{choices}\:{is}\:{possible}\:{if}\:{a} \\ $$$${student}\:{most}\:{answer}\:{to}\:\mathrm{3}\:{questions} \\ $$$${at}\:{least}? \\ $$$$\left.\mathrm{3}\right){How}\:{many}\:{choices}\:{is}\:{possible}\:{if}\:{a} \\ $$$${student}\:{most}\:{answer}\:{to}\:{at}\:{least}\:\mathrm{4} \\ $$$${questions}\:{of}\:\:{the}\:{five}\:{first}\:{questions}? \\ $$
Answered by mr W last updated on 07/Jul/20
1)  to select 8 from 10 questions there are  C_8 ^(10) =45 possibilities.  2)  do you mean 3 certain questions must  be selected? then one can select 5 from  the remaining 7 questions, there are  C_5 ^7 =21 ways.  3)  to select at least 4 from the first 5  questions, there are  C_4 ^5  ways, then select another 4 from  the remaining 6 questions, there are  C_4 ^6  ways, totally  C_4 ^5 ×C_4 ^6 =75 ways.
$$\left.\mathrm{1}\right) \\ $$$${to}\:{select}\:\mathrm{8}\:{from}\:\mathrm{10}\:{questions}\:{there}\:{are} \\ $$$${C}_{\mathrm{8}} ^{\mathrm{10}} =\mathrm{45}\:{possibilities}. \\ $$$$\left.\mathrm{2}\right) \\ $$$${do}\:{you}\:{mean}\:\mathrm{3}\:{certain}\:{questions}\:{must} \\ $$$${be}\:{selected}?\:{then}\:{one}\:{can}\:{select}\:\mathrm{5}\:{from} \\ $$$${the}\:{remaining}\:\mathrm{7}\:{questions},\:{there}\:{are} \\ $$$${C}_{\mathrm{5}} ^{\mathrm{7}} =\mathrm{21}\:{ways}. \\ $$$$\left.\mathrm{3}\right) \\ $$$${to}\:{select}\:{at}\:{least}\:\mathrm{4}\:{from}\:{the}\:{first}\:\mathrm{5} \\ $$$${questions},\:{there}\:{are} \\ $$$${C}_{\mathrm{4}} ^{\mathrm{5}} \:{ways},\:{then}\:{select}\:{another}\:\mathrm{4}\:{from} \\ $$$${the}\:{remaining}\:\mathrm{6}\:{questions},\:{there}\:{are} \\ $$$${C}_{\mathrm{4}} ^{\mathrm{6}} \:{ways},\:{totally} \\ $$$${C}_{\mathrm{4}} ^{\mathrm{5}} ×{C}_{\mathrm{4}} ^{\mathrm{6}} =\mathrm{75}\:{ways}. \\ $$

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