ANS = D.
we need to arrange - C=consonant, V=Vowel
we separate the C's from the V's to find diff. possible arrangements of vowels alone, and consonants alone. then we string them together as above. We get a set of (really) different arrangements of V's and a set of (really) different arrangements of C's and when we string one from each set together we get a (really) new and unique word.
Lets go -
For V's we have to CHOOSE where to put the A's (as they are all the same - we simply choose 3 locations out of the 5 for the A's).
That is 5C3.
then we have two options to place the I and the U must be in the last available place.
Total number of arrangements for V's - 5C3x2x1 = 10x2x1 = 20.
C's - more annoying - we CHOOSE the two locations for the P's 6C2 and then two locations for the T's 4C2 and then place the L in one of two places and the R lands in the last available place.
number of arrangements for C's - 6C2 x 4C2 x2x1 = 15x6x2x1 = 180
total words = 180x20 = 3600.
Further explanations - why can we separate C's from V's? we will eventually arrange them in the proper order as described above. IF THIS ORDER IS PREDETERMINED we can look at the problem as if it is two separate problems. There is no possibility of interchange between the C's and the V's as they must remain in the same relations one to another!!! we can't create a permutation where C's and V's switch places! So the problem of C's and problem of V's are separate problems.
then there is the counting issue - when we have three A's that are exactly the same (not in diff. colors for example
) then the order of V's - AAIUA is exactly the same as the order AAIUA (see what I mean - I switched the first two A's and you didn't even notice
So we start by deciding where the three places to put the A's are going to be. Each different selection of 3 places will give a different word - but we don't have to switch between the A's because that will leave us with exactly the same word).
If after all that the ans is still wrong
- well, I give up