SOLUTION: A MOTHER AND FATHER AND THEIR FOUR CHILDREN SIT AT A LONG BREAKFAST TABLE. THE TABLE IS PLACED AGAINST THE WALL, WITH ONE CHAIR AT EACH END AND FOUR CHAIRS ALONG THE EXPOSED SIDE.
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-> SOLUTION: A MOTHER AND FATHER AND THEIR FOUR CHILDREN SIT AT A LONG BREAKFAST TABLE. THE TABLE IS PLACED AGAINST THE WALL, WITH ONE CHAIR AT EACH END AND FOUR CHAIRS ALONG THE EXPOSED SIDE.
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Question 473705: A MOTHER AND FATHER AND THEIR FOUR CHILDREN SIT AT A LONG BREAKFAST TABLE. THE TABLE IS PLACED AGAINST THE WALL, WITH ONE CHAIR AT EACH END AND FOUR CHAIRS ALONG THE EXPOSED SIDE. HOW MANY WAYS CAN THE FAMILY BE ARRANGED IN SIX CHAIRS WITH THE PARENTS AT THE ENDS AND THE CHILDREN ALONG THE SIDE?
I understand this is probability however i'm not sure how to set it up, could someone show me how to set it up and work it out as well?
You can put this solution on YOUR website! the parents at the ends can be arranged in 2 ways.
mother on left and father on right.
mother on right and father on left.
the 4 spaces in the middle can be arranged 4! ways.
that's 4*3*2*1 = 24 ways.
2*24 = 48 ways total.
You can put this solution on YOUR website! A MOTHER AND FATHER AND THEIR FOUR CHILDREN SIT AT A LONG BREAKFAST TABLE. THE TABLE IS PLACED AGAINST THE WALL, WITH ONE CHAIR AT EACH END AND FOUR CHAIRS ALONG THE EXPOSED SIDE. HOW MANY WAYS CAN THE FAMILY BE ARRANGED IN SIX CHAIRS WITH THE PARENTS AT THE ENDS AND THE CHILDREN ALONG THE SIDE?
I understand this is probability however i'm not sure how to set it up, could someone show me how to set it up and work it out as well?
Choose someone to sit at A as either of 2 parents.
Choose someone to sit at B as the 1 remaining parent.
Choose someone to sit at C as any of 4 children.
Choose someone to sit at D as any of the remaining 3 children.
Choose someone to sit at E as either of the remaining 2 children.
Choose someone to sit at D as the 1 remaining child.
That's 2*1*4*3*2*1 = 48 ways.
Edwin
