Posted by Tuhin on Saturday, September 5, 2009 at 2:51pm.
1.
SEVENTEEN = EEEE NN S T V
So there is one group of 4 letters, one of 2 letters, and three of 1 letter for at total of 9 letters.
The number of permutations is
P(9,9)/(4!2!1!1!1!)
=9!/(4!2!1!1!1!)
2. Think of the two questions she can leave out. She has a choice of 12 for the first one, and 11 for the second. Thus she can make 12*11=132 different exams, if the order of the questions do not matter.
If the order of the questions matters, then we choose the ten questions one after another, which gives
12*11*10*9*8*7*6*5*4*3
=12!/2!
different exams.
3. How many choices do you have for choosing the first of three letters? (5)
How many for the second? (4)
How many for the third? (3)
How many arrangements in all?
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