Question 1018374
A photographer is lining up 5 family members in a line for a picture.
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Let the mother be M, father be F, and the children be A,B,C
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How many different lineups can there be if the mother and
father stand by side? 
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Number of permutations of 4 things {(MF),A,B,C}
plus
Number of permutations of 4 things {(FM),A,B,C} 

Answer: 4P4 + 4P4 = 4! + 4! = 2*4! = 2*24 = 48
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If the mother and father are on either end? 
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Number of permutations of MF{A,B,C}
plus
Number of permutations of FM{A,B,C}
plus
Number of permutations of {A,B,C}MF
plus
Number of permutations of {A,B,C}FM

Each is permutations of 3 things:

Answer = 4(3P3) = 4(3!) = 4(6)=24 ways
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The father is on the left and the mother is on the right?
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Number of permutations of F{A,B,C}M

Answer: 3P3 = 3! = 6 ways.

Edwin</pre>