Permutation Problem SWITZERLAND

[MichaelLiu]

Given the word "SWITZERLAND", in how many ways can we rearrange its letters so:

a) There is at least one consonant between every vowel

b) There is at least least two consonants between every vowel

Thanks for your help!

 

[anemone]

Hello MichaelLiu! Welcome to MHB!

For part (a), I would calculate the number of ways to rearrange the word "SWITZERLAND" such that there is no consonant between every vowel, and then find the total number of ways to rearrange all the letters in the word "SWITZERLAND". After that, do the subtraction to get the answer.

Total number of ways to rearrange all the letters in the word "SWITZERLAND" = $11! = 39916800$

There are 3 different possible cases to rearrange the word "SWITZERLAND" such that there is no consonant between every vowel.

Case 1 (The group of vowels (IEA) is positioned in the middle of the word) so $n_1=3!\cdot 7 \cdot 8!=1693440$

Case 2 (The group of vowels (IEA) is positioned at the beginning of the word) so $n_2=3!\cdot 8!=241920$

Case 3 (The group of vowels (IEA) is positioned at the end of the word) so $n_3=n_2=241920$

$\therefore n\text{(There is at least one consonant between every vowel)}=1693440-2(241920)=1209600$

 

[MichaelLiu]

Thanks a lot, @anemone !

Do you know the solution for part (b) as well?

 

[Opalg]

Given the word "SWITZERLAND", in how many ways can we rearrange its letters so:

a) There is at least one consonant between every vowel

b) There is at least least two consonants between every vowel

Thanks for your help!

I'm assuming that you want at least one consonant between each pair of vowels for part (a), and similarly in part (b).

There are 8 consonants (all different) and 3 vowels (all different) in SWITZERLAND. Remove the vowels and write the consonants in some order, with a gap between each pair of letters, and also gaps at the beginning and end, like this:

_ S _ W _ T _ Z _ R _ L _ N _ D _ . (Here, the consonants come in their natural order, but they could be in any of 8! possible orderings.)

For part (a), the vowels have to be inserted in the gaps, with at most one vowel in each gap. There are 9 gaps, so there are 9 places to put the I, then 8 places to put the E, and then 7 places to put the A. That gives a total of $8!\cdot9\cdot8\cdot7 = 20\,321\,280$ possible arrangements.

Part (b) is more complicated. Again, there are 8! possible orderings for the consonants. Write them with 9 gaps, as before, and insert the vowels. This time, we must avoid having vowels in adjacent gaps. I'll leave you to work out how many ways that can be done – I make it 35 possible ways for each of the 3! orderings of the vowels. That gives the answer for part (b) as $8!\cdot3!\cdot35 = 8\,467\,200.$