Treating each successive vowel in turn as if it were a consonant gives these five counts:
9145749347693960601600000
1959803431648705843200000
186834593817176623718400000
1959803431648705843200000
1959803431648705843200000
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 pseudovowel 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
201859753459816701849600000
The four numbers to be subtracted are:
16331695263739215600000
58794102949461176160000
58794102949461176160000
9799017158243529360000
What's left is
201716034541495796752320000

Posted by Charlie
on 20140926 14:44:57 