Below is a more complex version of the posted problem, along with an alternate approach.
A family and 5 children, including Adam and Bob, are standing in a line. The father wants to keep an eye on Adam and Bob and insists that Adam and Bob stand ahead of him at all times. How many ways can the family stand in the line such that the father is able to watch Adam and Bob?
Let the line proceed as follows:
Front...Back.
Let A = Adam, B = Bob, and F = father.
Let C, D and E = the other 3 children.
Number of options for C = 6. (Any of the 6 positions.)
Number of options for D = 5. (Any of the 5 remaining positions.)
Number of options for E = 4. (Any of the 4 remaining positions.)
Of the 3 remaining positions, the RIGHTMOST position must be occupied by F, so that he can keep an eye on Adam and Bob.
Thus:
Number of options for F = 1. (The rightmost remaining position.)
Number of options for A = 2. (Either of the 2 remaining positions.)
Number of options for B = 1. (Only 1 position left.).
To combine all of these options, we multiply:
6*5*4*1*2*1 = 240.
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