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Mary, Mary, Quite Orderly (Posted on 2014-09-24) Difficulty: 3 of 5
Determine the total number of ways the letters of the word supercalifragilisticexpialidocious can be arranged such that:

(i) All the five vowels will occur in order.
(ii) At least four of the five vowels will occur in a precisely reversed order.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution solution | Comment 1 of 6
SUPERCALIFRAGILISTICEXPIALIDOCIOUS contains:

3 a's v
0 b's
3 c's
1 d's
2 e's v
1 f's
1 g's
0 h's
7 i's v
0 j's
0 k's
3 l's
0 m's
0 n's
2 o's v
2 p's
0 q's
2 r's
3 s's
1 t's
2 u's v
0 v's
0 w's
1 x's
0 y's
0 z's

That's 16 vowels and 18 consonants.

For Part (i):

As far as vowels go, the order is given, so we're concerned only with which 16 positions out of the 34 are used by the vowels: C(34,16) =  2203961430.

In the remaining 18 positions, we are concerned with the order, but there are duplicates, within which we don't care which P comes first, for example. The 18 consonants have only ten identities, found in quantities of 3, 1, 1, 1, 3, 2, 2, 3, 1 and 1. So the consonants multiply the vowel result by 18! / (3! * 3! * 2 * 2 * 3!) =  7410154752000.

The answer for part (i) is then C(34,16) * 18! / (3! * 3! * 2 * 2 * 3!) = 2203961430 * 7410154752000 = 1633169526373921560000.

For Part (ii):

Calculation is done similarly to the above, but six cases need to be considered: all except A are in reverse order, all except E, etc, and all five are in reverse order. The cases are then added.

The simplest is that all five are in reverse order: that's the same as the answer to part (i) as it's still a specified order.

In the other cases, the vowel left out of the reverse order is treated just like a consonant.

A: C(34,13) * 21! / (3! * 3! * 3! * 2 * 2 * 3!) = 9145749347693960601600000

E: C(34,14) * 20! / (2 * 3! * 3! * 2 * 2 * 3!)

I: C(34,9) * 25! / (7! * 3! * 3! * 2 * 2 * 3!)

O: C(34,14) * 20! / (2 * 3! * 3! * 2 * 2 * 3!)

U: C(34,14) * 20! / (2 * 3! * 3! * 2 * 2 * 3!)

The six numbers to be added are:

    16331695263739215360000
  9145749347693960601600000
  1959803431648705843200000
186834593817176623718400000
  1959803431648705843200000
  1959803431648705843200000

The total is 201876085155080441064960000 to answer part (ii).

Note that in treating a given vowel as a consonant, it can be interspersed with the other vowels, which I assume is what is allowed by the rules. That is, it need not be in successive vowel positions, so that successive vowels might be, for example, ignoring the consonants: UUOAOIIIAIAIIIEE.


  Posted by Charlie on 2014-09-24 16:00:59
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