SOLUTION: Eight couples (husband and wife)are present at a meeting where a commitee of 3 is to be chesen . How many ways can this be done so that the commitee
a)Contains a couple? b)Ca
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-> SOLUTION: Eight couples (husband and wife)are present at a meeting where a commitee of 3 is to be chesen . How many ways can this be done so that the commitee
a)Contains a couple? b)Ca
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Question 34706This question is from textbook Finite Math
: Eight couples (husband and wife)are present at a meeting where a commitee of 3 is to be chesen . How many ways can this be done so that the commitee
a)Contains a couple? b)Cantains no couple? This question is from textbook Finite Math
You can put this solution on YOUR website! a) contains a couple 8 choices
Contains one other person 14
# of ways to choose the committee with one couple = 8(14)=112 ways
b) contains no couple
Choose a person 16 ways
Choose a 2nd person 14 ways (spouse of the 1st choice cannot be chosen)
Choose a 3rd person 12 ways (spouse of the 1st and 2nd cannot be chosen)
# of ways to choose thecommittee with no couple = 16*14*12=2688 ways.
Cheers,
Stan H.
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