Treating each successive vowel in turn as if it were a consonant gives these five counts:
however, when added together, each case where two successive letters (A & E, E & I, I & O, and O & U) are together, possibly interspersed, possibly switched, possibly correctly ordered, has been counted twice, with some of the more ordered ones counted more than twice.
So we have to subtract out one set of all cases where A&E all appear as successive vowels, mixed up or not, without any other intervening vowels, then E&I, etc.
Where do the part (i) orders fall in this. They're supposed to be counted once. We've added them in 5 times and subtracted them out 4 times. That sounds correct. So once we've done the above additions and subtractions we should be done.
The pseudo-vowel counts:
AE: 5 = 3 + 2
EI: 9 = 2 + 7
IO: 9 = 7 + 2
OU: 4 = 2 + 2
The completely reversed count was 1633169526373921560000. If we mix up in some order or not we get for:
AE: 1633169526373921560000 * 5! / (3! * 2)
EI: 1633169526373921560000 * 9! / (2 * 7!)
IO: 1633169526373921560000 * 9! / (7! * 2)
OU: 1633169526373921560000 * 4! / (2 * 2)
The five individual vowel totals add up to
The four numbers to be subtracted are:
What's left is
Posted by Charlie
on 2014-09-26 14:44:57