Posted by Shayan Siddiqi on Thursday, September 23, 2010 at 7:39pm.
You must be familiar with the usual conventions of permutations, P(i,j), the number of ways of ordering i items out of j distinct items, where
P(i,j)=j!/(j-i)!
For example, how many 2 lettered words can be made of the letters a,b and c:
P(2,3)=3!/(3-2)!=3!/1!=6
a)
You would be making 5 lettered words out of the 5 letters STEVN or STVIN.
The number of words:
P(5,5)+P(5,5)
b)
You'd be making words out of
STEVI, STEIN, SEVIN, TEVIN
for a total of
4P(5,5) words.
However, half of these words have the vowels in wrong order, so the answer should be 2P(5,5).
c)
Treat EI as one letter X, and choose 3 remaining letters to give
STVX, STNX, SVNX, TVNX
So you have 4P(4,4) words.
d)
The two cases are mutually exclusive, so we just have to add cases a) and c).
Check my work.