Question 1098700
Ans:  {{{ highlight(36) }}} different course codes
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Workout:
There are 3 unique arrangements of MAA (to see this, assume you had {{{ M }}}, {{{A[1]}}}, and {{{A[2] }}}, there would be 3! = 6 ways to arrange those, but we need to divide by 2!,  because {{{A[1] }}} and {{{A[2] }}} are really just "A" and therefore not distinguishable, so we need to divide that 6 by the number of arrangements of {{{ A[1] }}} and {{{ A[2] }}}:  6/2 = 3)
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There are 12 unique arrangements of 2,3,1,2  (4!/2! = 24/2 = 12)
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3*12 = 36