Li0n
spiKu
*hugs*ND said:Here's my solution: (btw, what exactly does "by inspecion" mean?)
Let C denote a consonant and V a vowel. First lay out the Cs:
C C C C C C C
There are now 8 gaps for the Vs, and since there are 4 Vs there are 8C4 ways of distributing the 4 Vs into the 8 gaps. So if all the Cs were identical and all the Vs were identical, there would be 8C4 ways of arranging them. But since there are 7 consonants, with 2 pairs being identical, there are 7!/(2!*2!) ways of arranging them, and since the two E's are indistinguishable, there is 1 way of arranging the vowels. So the answer is:
8C4*7!/(2!*2!)*1 = 88200
i thought inspection also meant writing out all the possible combinations, in your head though...